ip地址有效需要满足：遍历完s，将其分为4段，每段数字有效性：范围在[0,255],且不含前导零
为避免重复取值，需要startIndex作为下次递归的初始位置。当遍历结束startIndex == len(s)，且ip段数为4 len(path) == 4加入结果，否则提前回溯。
如果当前字符串满足要求，递归调用下一层

func restoreIpAddresses(s string) []string {res := []string{}path := []string{}var help func(s string, startIndex int)help = func(s string, startIndex int) {if len(path) == 4 {if startIndex == len(s) {res = append(res, strings.Join(path, "."))}return}for i := startIndex; i < len(s); i++ {str := s[startIndex : i+1]if isValid(str) {path = append(path, str)help(s, i+1)path = path[:len(path)-1]} else {break}}}help(s, 0)return res
}func isValid(str string) bool {if len(str) != 1 && str[0] == '0' {return false}i, _ := strconv.Atoi(str)if i > 255 {return false}return true
}

子集相当于找树的节点，只要遍历到一个节点就加入结果

func subsets(nums []int) [][]int {res := [][]int{}path := []int{}var help func(nums []int, startIndex int)help = func(nums []int, startIndex int) {tmp := make([]int,len(path))copy(tmp,path)res = append(res, tmp)for i := startIndex; i < len(nums); i++ {path = append(path, nums[i])help(nums, i+1)path = path[:len(path)-1]}}help(nums, 0)return res
}

相较于上题子集问题，数组有重复元素，要求结果集合不能重复，需要去重
先对数组进行排序，然后同层去重【否则，相当于当前元素要走前一个元素已经走过的路】
同层去重：used[i - 1] = false,说明是从前一个元素回溯回来的，若used[i - 1] = true说明是进入下一个递归取值

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