Huangzj
72 lines
1.6 KiB
Go
72 lines
1.6 KiB
Go
/*
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* @Author : huangzj
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* @Time : 2021/3/23 17:06
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* @Description：
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*/
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package FullPermutation
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func DigitDecreases(b []int) [][]int {
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result := make([][]int, 0)
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bit := make([]int, len(b)-1)
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sum := 1
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//计算中的结果数,阶乘
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for i := 1; i <= len(b); i++ {
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sum = sum * i
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}
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for i := 0; i < sum; i++ {
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result = append(result, getRangeBySpace1(b, bit))
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bit = decreaseBit(bit)
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}
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return result
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}
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//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
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//origin 原始数组
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//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
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func getRangeBySpace1(origin []int, increaseBit []int) []int {
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permutation := make([]int, len(origin))
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flag := make(map[int]bool, 0)
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//排列数组从左往右，即从大到小
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for i := 0; i < len(origin)-1; i++ {
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p := increaseBit[i] //从右到左，对应数字应该填充的位置
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j := len(origin) - 1 //该数字在本次排列的位置
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for ; ; j-- {
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//该位置未被填充
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if !flag[j] {
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p--
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}
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//找到对应位置需要推出循环
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if p < 0 {
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flag[j] = true
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break
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}
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}
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permutation[j] = origin[i]
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}
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for i := 0; i < len(origin); i++ {
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if !flag[i] {
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permutation[i] = origin[len(origin)-1]
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}
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}
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return permutation
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}
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//递减进制位
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//每次加一,计算新的数组结果
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func decreaseBit(bit []int) []int {
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//数组下标从小到大,进制位从大到小
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for i := 0; i < len(bit); i++ {
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if bit[i]+1 >= len(bit)+1-i {
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bit[i] = (bit[i] + 1) % (len(bit) + 1 - i)
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} else {
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bit[i] += 1
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break
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}
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}
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return bit
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}
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