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IP
        if (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;
    }
};