54 lines
1.2 KiB
Go
54 lines
1.2 KiB
Go
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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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