Huangzj
72 lines
1.6 KiB
Go
72 lines
1.6 KiB
Go
/*
|
||
* @Author : huangzj
|
||
* @Time : 2021/3/23 17:05
|
||
* @Description：
|
||
*/
|
||
|
||
package FullPermutation
|
||
|
||
func DigitIncrease(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, getRangeBySpace(b, bit))
|
||
bit = increaseBit(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 getRangeBySpace(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 := len(origin) - 1; i > 0; i-- {
|
||
p := increaseBit[i-1] //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
|
||
//中介数计算从左到右，即从大到小
|
||
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[0]
|
||
}
|
||
}
|
||
|
||
return permutation
|
||
}
|
||
|
||
//递增进制位
|
||
//根据递增进制位进行+1操作，返回递增的结果
|
||
func increaseBit(bit []int) []int {
|
||
//最开始是从二进制开始
|
||
for i, num := range bit {
|
||
if num+1 >= i+2 {
|
||
bit[i] = (num + 1) % (i + 2)
|
||
} else {
|
||
bit[i] = num + 1
|
||
break
|
||
}
|
||
}
|
||
|
||
return bit
|
||
}
|