feat(Go-Tool):2021/03/26: 新增全排列的递归实现、字典序实现、递增进制位实现、递减进制位实现

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Huangzj
2021-03-26 09:13:34 +08:00
parent 00ada130fa
commit 092ab58276
10 changed files with 479 additions and 0 deletions
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/*
* @Author : huangzj
* @Time : 2021/3/23 17:06
* @Description
*/
package FullPermutation
func DigitDecreases(b []int) [][]int {
result := make([][]int, 0)
bit := make([]int, len(b)-1)
sum := 1
//计算中的结果数,阶乘
for i := 1; i <= len(b); i++ {
sum = sum * i
}
for i := 0; i < sum; i++ {
result = append(result, getRangeBySpace1(b, bit))
bit = decreaseBit(bit)
}
return result
}
//根据递增进制位,并且根据【数空格法】在原来的数组中求得全排列的一种可能实现
//origin 原始数组
//increaseBit 递增进制位,比原始数组长度小1,这里通过数组实现,更简单
func getRangeBySpace1(origin []int, increaseBit []int) []int {
permutation := make([]int, len(origin))
flag := make(map[int]bool, 0)
//排列数组从左往右,即从大到小
for i := 0; i < len(origin)-1; i++ {
p := increaseBit[i] //从右到左,对应数字应该填充的位置
j := len(origin) - 1 //该数字在本次排列的位置
for ; ; j-- {
//该位置未被填充
if !flag[j] {
p--
}
//找到对应位置需要推出循环
if p < 0 {
flag[j] = true
break
}
}
permutation[j] = origin[i]
}
for i := 0; i < len(origin); i++ {
if !flag[i] {
permutation[i] = origin[len(origin)-1]
}
}
return permutation
}
//递减进制位
//每次加一,计算新的数组结果
func decreaseBit(bit []int) []int {
//数组下标从小到大,进制位从大到小
for i := 0; i < len(bit); i++ {
if bit[i]+1 >= len(bit)+1-i {
bit[i] = (bit[i] + 1) % (len(bit) + 1 - i)
} else {
bit[i] += 1
break
}
}
return bit
}