切割问题可以使用回溯，本题分别两步，切割字符串和判断IP
切割逻辑如下：递归用来纵向遍历，for循环用来横向遍历，切割线（就是图中的红线）切割到字符串的结尾位置

代码如下

class Solution {
private:vector<string> result;void backtracking(string&s, int startIndex, int pointNum) {if (pointNum == 3) {if (isValid(s, startIndex, s.size() - 1 )) {result.push_back(s);}}for (int i = startIndex; i < s.size(); i++) {if(isValid(s, startIndex, i)) {s.insert(s.begin() + i + 1 , '.');pointNum++;backtracking(s, i + 2, pointNum);pointNum--;s.erase(s.begin() +i + 1);}else break;}}bool isValid (string& s, int start, int end) {//判断有效IPif (start > end) return false;if (s[start] == '0' && start != end) return false;int num = 0;for (int i = start; i <= end; i++) {if (s[i] > '9' || s[i] < '0') return false;num = num * 10 + s[i] - '0';if (num > 255) return false; }return true;} 
public:vector<string> restoreIpAddresses(string s) {result.clear();if (s.size() < 4 || s.size() > 12) return result; // 剪枝了backtracking(s, 0, 0);return result;}
};

分割问题找树的叶子节点
回溯法

代码如下

class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& nums, int startIndex) {result.push_back(path); if (startIndex > nums.size()) {//startIndex已经大于数组的长度为终止return;}for (int i = startIndex; i < nums.size(); i++) {path.push_back(nums[i]);backtracking(nums, i + 1);path.pop_back();}}
public:vector<vector<int>> subsets(vector<int>& nums) {result.clear();path.clear();  backtracking(nums, 0);return result;}
};

只比78.子集多了一个去重，我们要去掉同一树层的相同元素，保留同一树枝的。如下图

代码如下

class Solution {
private:vector<vector<int>> result;vector<int> path;void backtracking(vector<int>& nums, int startIndex, vector<bool>& used) {result.push_back(path);if (startIndex > nums.size()) {return;}for (int i = startIndex; i < nums.size(); i++) {//同一树枝可重复取，但同一树层不可if(i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {continue;}path.push_back(nums[i]);used[i] = true;backtracking(nums, i + 1, used);used[i] = false;path.pop_back();}}
public:vector<vector<int>> subsetsWithDup(vector<int>& nums) {result.clear();path.clear();  vector<bool> used(nums.size(), false);backtracking(nums, 0, used);return result;}
};