feat(Go-Tool)：2021/03/18：新增超过半数数字系列算法题，新增出现次数不同的数字系列算法题
This commit is contained in:
Huangzj
@@ -0,0 +1,11 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 16:46
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* @Description：
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*/
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package numberList
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type i interface {
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Doc() string
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}
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@@ -0,0 +1,48 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/18 13:08
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* @Description：
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*/
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package NumberCount
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type OnlyOneNumberOtherKObj struct{}
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func (*OnlyOneNumberOtherKObj) Doc() string {
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return `
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给定一个整型数组 arr和一个大于1的整数k。已知 arr中只有1个数出现了1次，其他的数都出现了k次，请返回只出现了1次的数。
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【要求】
时间复杂度为 O（N），额外空间复杂度为 O（1）。
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`
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}
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func OnlyOneNumberOtherK(numList []int, k int) int {
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bitList := make([]int, 32)
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for _, num := range numList {
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//把每个数字转换成二进制
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store := make([]int, 32)
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for i := 0; num != 0; i++ {
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store[i] = num % k
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num = num / k
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}
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for j := 0; j < 32; j++ {
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//不进位加法
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bitList[j] = (store[j] + bitList[j]) % k
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}
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}
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//k进制转换回十进制
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power := 1 //k的次方值
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result := 0 //只出现一次的数字
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for i, bit := range bitList {
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power = 1
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for j := 0; j < i; j++ {
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power = power * k
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}
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result = result + power*bit
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}
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return result
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}
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@@ -0,0 +1,25 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/18 13:09
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* @Description：
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*/
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package NumberCount
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import (
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"fmt"
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"testing"
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)
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func TestOnlyOneNumberOtherK(t *testing.T) {
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fmt.Println(OnlyOneNumberOtherK([]int{1, 2, 3, 4, 2, 3, 4}, 2)) //1
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fmt.Println(OnlyOneNumberOtherK([]int{1, 2, 3, 4, 2, 3, 4, 2, 3, 4}, 3)) //1
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fmt.Println(OnlyOneNumberOtherK([]int{1, 2, 3, 4, 4, 5, 6, 2, 2, 3, 5, 5, 3, 4, 6, 6}, 3)) //1
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fmt.Println(OnlyOneNumberOtherK([]int{1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}, 10)) //1
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fmt.Println(OnlyOneNumberOtherK([]int{2, 3, 3000, 4, 2, 3, 4}, 2)) //3000
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fmt.Println(OnlyOneNumberOtherK([]int{2, 3, 4, 2, 9999, 3, 4, 2, 3, 4}, 3)) //9999
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fmt.Println(OnlyOneNumberOtherK([]int{2, 3, 4, 4, 5, 6, 321, 2, 2, 3, 5, 5, 3, 4, 6, 6}, 3)) //321
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fmt.Println(OnlyOneNumberOtherK([]int{2, 2, 2, 2, 2, 2, 2, 5891, 2, 2, 2}, 10)) //5891
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}
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@@ -0,0 +1,53 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 21:53
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* @Description：
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*/
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package NumberCount
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type OnlyOneNumberOtherThreeObj struct{}
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func (*OnlyOneNumberOtherThreeObj) Doc() string {
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return `
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给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。
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要求：线性时间复杂度。 不使用额外空间来实现
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`
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}
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func OnlyOneNumberOtherThree1(numList []int) int {
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bitList := make([]int, 32)
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for _, num := range numList {
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for i := 0; i < 32; i++ {
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bitList[i] += (num >> i) & 1
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}
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}
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res := 0
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for i := 0; i < 32; i++ {
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if bitList[i]%3 != 0 {
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res += 1 << i
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}
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}
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return res
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}
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//通过二进制的方式，本质上就是要构造一个相应的回路，即出现num的次数分别从1~3应该是 01 -> 10 -> 00 ...
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func OnlyOneNumberOtherThree2(numList []int) int {
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var a uint
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var b uint
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//第一次出现num计算的结果 a = num ,b = 0
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//第二次出现num计算的结果 a = 0 ,b = num
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//第三次出现num计算的结果 a = num ,b = 0
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//因此直接返回a即可
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for _, num := range numList {
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a = a ^ uint(num)&(^b)
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b = b ^ uint(num)&(^a)
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}
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return int(a)
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}
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@@ -0,0 +1,26 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 21:55
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* @Description：
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*/
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package NumberCount
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import (
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"fmt"
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"testing"
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)
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func TestOnlyOneNumberOtherThree1(t *testing.T) {
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fmt.Println(OnlyOneNumberOtherThree1([]int{1, 1, 1, 2, 2, 2, 3, 3, 3, 4})) //4
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fmt.Println(OnlyOneNumberOtherThree1([]int{1, 2, 3, 1, 2, 3, 1, 2, 3, 4})) //4
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fmt.Println(OnlyOneNumberOtherThree1([]int{1, 15, 23, 67, 15, 67, 67, 23, 23, 14, 15, 1, 1})) //14
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fmt.Println(OnlyOneNumberOtherThree1([]int{100, 200, 300, 4, 6, 100, 100, 6, 6, 4, 200, 300, 200, 300, 5, 4})) //5
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}
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func TestOnlyOneNumberOtherThree2(t *testing.T) {
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fmt.Println(OnlyOneNumberOtherThree2([]int{1, 1, 1, 2, 2, 2, 3, 3, 3, 4})) //4
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fmt.Println(OnlyOneNumberOtherThree2([]int{1, 2, 3, 1, 2, 3, 1, 2, 3, 4})) //4
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fmt.Println(OnlyOneNumberOtherThree2([]int{1, 15, 23, 67, 15, 67, 67, 23, 23, 14, 15, 1, 1})) //14
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fmt.Println(OnlyOneNumberOtherThree2([]int{100, 200, 300, 4, 6, 100, 100, 6, 6, 4, 200, 300, 200, 300, 5, 4})) //5
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}
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@@ -0,0 +1,25 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 17:51
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* @Description：
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*/
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package NumberCount
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type OnlyOneNumberShowOnceObj struct{}
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func (*OnlyOneNumberShowOnceObj) Doc() string {
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return `
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一个整型数组里除了一个数字之外，其他的数字都出现了两次。请写程序找出这两个只出现一次的数字。要求时间复杂度为O(n)，空间复杂度为O(1)。
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`
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}
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//通过异或解决
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func OnlyOneNumberShowOnce(numberList []int) int {
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result := 0
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for _, num := range numberList {
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result ^= num
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}
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return result
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}
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@@ -0,0 +1,19 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 17:52
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* @Description：
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*/
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package NumberCount
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import (
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"fmt"
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"testing"
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)
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func TestOnlyOneNumberShowOnce(t *testing.T) {
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fmt.Println(OnlyOneNumberShowOnce([]int{1, 2, 3, 2, 3, 4, 4})) //1
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fmt.Println(OnlyOneNumberShowOnce([]int{1, 2, 3, 2, 3, 4, 4, 1, 5})) //5
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fmt.Println(OnlyOneNumberShowOnce([]int{1, 2, 3, 2, 3, 4, 4, 1, 10})) //10
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fmt.Println(OnlyOneNumberShowOnce([]int{2, 2, 2, 2, 1, 3, 3, 3, 3})) //1
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}
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@@ -0,0 +1,58 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 18:01
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* @Description：
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*/
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package NumberCount
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type OnlyTwoNumberShowOnceObj struct {
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}
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func (*OnlyTwoNumberShowOnceObj) Doc() string {
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return `
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一个整型数组里除了两个数字之外，其他的数字都出现了两次。请写程序找出这两个只出现一次的数字。要求时间复杂度为O(n)，空间复杂度为O(1)。
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`
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}
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func OnlyTwoNumberShowOnce(numList []int) (int, int) {
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if len(numList) < 2 {
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panic("数组长度不能小于两个")
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}
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result := 0
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//先进行异或，得到异或的结果
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for _, num := range numList {
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result = result ^ num
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}
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//找到异或后为1的位置
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pos := 0
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for ; pos < 32; pos++ {
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if ((result >> pos) & 1) == 1 {
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break
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}
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}
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firstList := make([]int, 0)
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secondList := make([]int, 0)
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//根据异或位的结果，把数组分成两组
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for _, num := range numList {
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if ((num >> pos) & 1) == 1 {
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firstList = append(firstList, num)
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} else {
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secondList = append(secondList, num)
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}
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}
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first := 0
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for _, num := range firstList {
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first ^= num
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}
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second := 0
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for _, num := range secondList {
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second ^= num
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}
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return first, second
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}
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@@ -0,0 +1,19 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 18:02
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* @Description：
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*/
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package NumberCount
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import (
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"fmt"
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"testing"
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)
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func TestOnlyTwoNumberShowOnce(t *testing.T) {
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fmt.Println(OnlyTwoNumberShowOnce([]int{2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 1, 7})) //1,7
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fmt.Println(OnlyTwoNumberShowOnce([]int{2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 1, 7, 8, 9, 0, 19, 19, 0, 8, 9})) //1,7
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fmt.Println(OnlyTwoNumberShowOnce([]int{2, 3, 4, 5, 1, 7, 5, 4, 3, 2})) //1,7
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fmt.Println(OnlyTwoNumberShowOnce([]int{2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 1, 7, 1, 100})) //1,100
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}
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@@ -0,0 +1,48 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 15:45
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* @Description：
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*/
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package NumberMoreThan
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type NumberMoreThanHalfObj struct{}
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func (n *NumberMoreThanHalfObj) Doc() string {
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return `
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题目：
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数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。
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例如输入一个长度为9的数组{1,2,3,2,2,2,5,4,2}。由于数字2在数组中出现了5次，超过数组长度的一半，因此输出2。
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要求：时间复杂度O(N),空间复杂度O(1)
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`
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}
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func NumberMoreThanHalf(numList []int) int {
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var number, count int
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for _, j := range numList {
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if number == j {
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count++
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} else if number != j && count <= 1 {
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number = j
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} else {
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//number !=j && count >1
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count--
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}
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}
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//最后验证，确保一定超过
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countValid := 0
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for _, j := range numList {
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if j == number {
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countValid++
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}
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}
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if countValid <= len(numList)/2 {
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return -1
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}
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return number
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}
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@@ -0,0 +1,20 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 16:23
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* @Description：
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*/
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package NumberMoreThan
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import (
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"fmt"
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"testing"
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)
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func TestNumberMoreThanHalf(t *testing.T) {
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fmt.Println(NumberMoreThanHalf([]int{1, 2, 3})) //-1
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fmt.Println(NumberMoreThanHalf([]int{1, 2, 3, 1, 1})) //1
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fmt.Println(NumberMoreThanHalf([]int{1, 2, 3, 1})) //-1
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fmt.Println(NumberMoreThanHalf([]int{1, 2, 3, 2, 3, 2, 3, 3})) //-1
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fmt.Println(NumberMoreThanHalf([]int{1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1})) //1
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}
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@@ -0,0 +1,68 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 16:46
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* @Description：
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*/
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package NumberMoreThan
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type NumberMoreThanKObj struct{}
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func (*NumberMoreThanKObj) Doc() string {
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return `
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给定一个整型数组arr(数组长度为N) 与一个整数k，打印所有出现次数大于 N/K 的数。如果没有这样的数，返回-1。
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要求：时间复杂度为O（N*K）,额外空间复杂度为O（K）。
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`
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}
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func NumberMoreThanK(numList []int, k int) []int {
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//小于2的情况是肯定不会超过.
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if k < 2 {
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return []int{}
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}
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//用来保存数字与其出现次数
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numMap := make(map[int]int, 0)
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for _, num := range numList {
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//如果不存在则数值设置为1
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if _, ok := numMap[num]; !ok {
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numMap[num] = 1
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} else {
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//存在则出现次数+1
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numMap[num] = numMap[num] + 1
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}
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//当容器的大小为k
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if len(numMap) == k {
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for key, value := range numMap {
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//个数正好等于1的要删掉，因为本次减一之后，就等于0了
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if value == 1 {
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delete(numMap, key)
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}
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}
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continue
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}
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}
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//map的key是可能出现次数超过N/k的数
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//这里主要是为了拿到key
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for key := range numMap {
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numMap[key] = 0
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}
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//重新计算对应数字的出现次数
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for _, num := range numList {
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if _, ok := numMap[num]; ok {
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numMap[num]++
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}
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}
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//出现次数超过k次的判断
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result := make([]int, 0)
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for key, value := range numMap {
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if value > len(numList)/k {
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result = append(result, key)
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}
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}
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return result
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}
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@@ -0,0 +1,19 @@
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/*
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* @Author : huangzj
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* @Time : 2021/3/16 16:59
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* @Description：
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*/
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package NumberMoreThan
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import (
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"fmt"
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"testing"
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)
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func TestNumberMoreThanK(t *testing.T) {
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fmt.Println(NumberMoreThanK([]int{1, 2, 3, 1, 2, 3, 1, 2, 3, 1}, 3)) //1
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fmt.Println(NumberMoreThanK([]int{1, 2, 3, 1, 2, 3, 1, 2, 3, 1}, 4)) //1,2,3
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fmt.Println(NumberMoreThanK([]int{1, 2, 3, 1, 2, 3, 1, 2, 3}, 2)) //
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fmt.Println(NumberMoreThanK([]int{1, 2, 3}, 2)) //
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}
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Reference in New Issue
Block a user