算法训练营day25(补)，回溯5
package main
import "sort"
491. 非递减子序列
func findSubsequences(nums []int) [][]int {
//存储全部集合
result := make([][]int, 0)
if len(nums) == 0 {
return result
}
//存储单次集合
path := make([]int, 0)
var backtrace func(numList []int, startIndex int)
backtrace = func(numList []int, startIndex int) {
if len(path) > 1 {
temp := make([]int, len(path))
copy(temp, path)
result = append(result, temp)
}
//记录数组每一个元素是否使用过
user := make(map[int]bool, len(nums))
for i := startIndex; i < len(numList); i++ {
if user[numList[i]] {
continue
}
if len(path) == 0 || numList[i] >= path[len(path)-1] {
path = append(path, numList[i])
user[numList[i]] = true
backtrace(numList, i+1)
//回溯处理
path = path[:len(path)-1]
}
}
}
backtrace(nums, 0)
return result
}
46. 全排列
func permute(nums []int) [][]int {
//存储全部集合
result := make([][]int, 0)
if len(nums) == 0 {
return result
}
//存储单次集合
path := make([]int, 0)
//记录数组每一个元素是否使用过
user := make(map[int]bool, len(nums))
var backtrace func(numList []int, user map[int]bool)
backtrace = func(numList []int, user map[int]bool) {
if len(path) == len(numList) {
temp := make([]int, len(path))
copy(temp, path)
result = append(result, temp)
}
//因为是组合所有元素又要
for i := 0; i < len(numList); i++ {
if user[numList[i]] { //已用过的数直接跳过
continue
}
path = append(path, numList[i])
user[numList[i]] = true
backtrace(numList, user)
//回溯处理
user[numList[i]] = false
path = path[:len(path)-1]
}
}
backtrace(nums, user)
return result
}
47. 全排列 II
func permuteUnique(nums []int) [][]int {
//存储全部集合
result := make([][]int, 0)
if len(nums) == 0 {
return result
}
sort.Ints(nums)
//存储单次集合
path := make([]int, 0)
//记录数组每一个元素是否使用过
user := make(map[int]bool, len(nums))
var backtrace func(numList []int, user map[int]bool)
backtrace = func(numList []int, user map[int]bool) {
if len(path) == len(numList) {
temp := make([]int, len(path))
copy(temp, path)
result = append(result, temp)
}
for i := 0; i < len(numList); i++ {
if i > 0 && numList[i] == numList[i-1] && user[i-1] == false { //过滤重复
continue
}
if user[i] { //已用过的数直接跳过
continue
}
path = append(path, numList[i])
user[i] = true
backtrace(numList, user)
//回溯处理
user[i] = false
path = path[:len(path)-1]
}
}
backtrace(nums, user)
return result
}