Leetcode 93-复原 IP 地址

题目描述

https://leetcode.cn/problems/restore-ip-addresses/description/

解题思路

这是一道切割问题，可以参考分割回文串的思路

采用 isValid 函数判断当前切割得到的子串是否有效

class Solution {
public:
    vector<string> res;
    bool isValid(string s, int start, int end){
        if(start > end) return false;
        if (s[start] == '0' && end - start > 0) return false;
        if (end - start + 1 > 3) return false; // 添加长度检查
        string sub = s.substr(start, end-start+1);
        int num = stoi(sub);
        if (num > 255) return false;
        return true;
    }
    void backTracking(string s, int startIndex, int dotNum){
        if(dotNum == 3){
            if (isValid(s, startIndex,s.size()-1)){
                res.push_back(s);
            }
            return;
        }
        for (int i = startIndex; i< startIndex+3 && i <s.size();i++){
            if (isValid(s, startIndex,i)){
                s.insert(s.begin()+i+1,'.');
                dotNum += 1;
                backTracking(s, i + 2, dotNum);
                dotNum -= 1;
                s.erase(s.begin()+i+1);
            }
        }
    }
    vector<string> restoreIpAddresses(string s) {
        backTracking(s,0,0);
        return res;
    }
};

Leetcode 78-子集

题目描述

https://leetcode.cn/problems/subsets/description/

解题思路

在子集问题中需要在每一层递归中进行结果的收集

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backTracking(vector<int>& nums, int count, int startIndex){
        if (path.size()==count) {
            res.push_back(path);
            return;
        }
        for (int i = startIndex; i <nums.size();i++){
            path.push_back(nums[i]);
            backTracking(nums,count, i+1);
            path.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        for (int i =0; i <=nums.size(); i++){
            backTracking(nums,i,0);
        }
        return res;
    }
};

Leetcode 90-子集 Ⅱ

题目描述

https://leetcode.cn/problems/subsets-ii/description/

解题思路

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backTracking(vector<int>& nums, int startIndex){
        res.push_back(path);
        if (startIndex == nums.size()) return;
        for (int i =startIndex; i < nums.size();i++){
            if (i >startIndex && nums[i]==nums[i-1]) continue;
            path.push_back(nums[i]);
            backTracking(nums,i+1);
            path.pop_back();
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        backTracking(nums,0);
        return res;
    }
};