同层去重，只需保证当前层元素不重复即可【前仆后继的感觉】

func findSubsequences(nums []int) [][]int {res, path := [][]int{}, []int{}var help func(nums []int, startIndex int)help = func(nums []int, startIndex int) {if len(path) > 1 {tmp := make([]int, len(path))copy(tmp, path)res = append(res, tmp)}used := make(map[int]bool, len(nums))for i := startIndex; i < len(nums); i++ {if used[nums[i]]|| len(path) > 0 && nums[i] < path[len(path)-1] {continue}path = append(path, nums[i])used[nums[i]] = truehelp(nums, i+1)path = path[:len(path)-1]}}help(nums, 0)return res
}

used数组，记录当前元素是否在path路径中存在，若存在则跳过。一个值在一条path路径中只能被选取一次

func permute(nums []int) [][]int {res := [][]int{}path := []int{}used := make([]bool, len(nums))var help func(nums []int, index int)help = func(nums []int, index int) {if len(path) == len(nums) {tmp := make([]int, len(path))copy(tmp, path)res = append(res, tmp)}for i := 0; i < len(nums); i++ {if used[i] == true {continue}path = append(path, nums[i])used[i] = truehelp(nums, i+1)path = path[:len(path)-1]used[i] = false}}help(nums, 0)return res
}

上题扩展，有重复元素。先排序，然后同层剪枝
同层剪枝：i > 0 && nums[i] == nums[i-1] && used[i-1] == false跳过，
当nums[i-1]作为某一层元素，其所有的递归选择已经穷尽，回溯时才会有used[i - 1]==false。因此，当nums[i] == nums[i-1]跳过，否则nums[i]递归得到的结果与nums[i - 1]结果重复

func permuteUnique(nums []int) [][]int {res := [][]int{}path := []int{}used := make([]bool, len(nums))sort.Ints(nums)var help func(nums []int, index int)help = func(nums []int, index int) {if len(path) == len(nums) {tmp := make([]int, len(path))copy(tmp, path)res = append(res, tmp)return}for i := 0; i < len(nums); i++ {if i > 0 && nums[i] == nums[i-1] && used[i-1] == false || used[i] == true {continue}used[i] = truepath = append(path, nums[i])help(nums, i+1)path = path[:len(path)-1]used[i] = false}}help(nums, 0)return res
}