LeetCode93. Restore IP Addresses

文章目录

一、题目

A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.

For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245", "192.168.1.312" and "192.168@1.1" are invalid IP addresses.

Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.

Example 1:

Input: s = "25525511135"

Output: "255.255.11.135","255.255.111.35"

Example 2:

Input: s = "0000"

Output: "0.0.0.0"

Example 3:

Input: s = "101023"

Output: "1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"

Constraints:

1 <= s.length <= 20

s consists of digits only.

二、题解

注意c++中字符串的insert方法和erase方法

cpp 复制代码
class Solution {
public:
    vector<string> res;
    bool isValid(string& s,int start,int end){
        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] < '0' || s[i] > '9') return false;
            num = num * 10 + s[i] - '0';
            if(num > 255) return false;
        }
        return true;
    }
    void backtracking(string s,int startIndex,int pointSum){
        if(pointSum == 3){
            if(isValid(s,startIndex,s.size()-1)){
                res.push_back(s);
                return;
            }
        }
        for(int i = startIndex;i < s.size();i++){
            //合法的情况下
            if(isValid(s,startIndex,i)){
                s.insert(s.begin() + i + 1,'.');
                pointSum++;
                backtracking(s,i + 2,pointSum);
                s.erase(s.begin() + i + 1);
                pointSum--;
            }
            else break;
        }
    }
    vector<string> restoreIpAddresses(string s) {
        backtracking(s,0,0);
        return res;
    }
};
相关推荐
To_OC1 天前
LC 1 两数之和:面试第一道必考题,暴力解法直接被面试官 pass
javascript·算法·leetcode
鱼鱼不愚与2 天前
《原来如此 | 第01期:为什么导航软件能预测红绿灯倒计时?》
算法
博客18002 天前
酷宝的使用方法,超好用的免费界面库,C++、MFC可用
c++·mfc·界面库·库来帮·酷宝
郝学胜_神的一滴2 天前
CMake 026:属性体系精讲、四大作用域全解 & 实战代码落地
c++·cmake
复杂网络2 天前
论最小 Agent 计算机的形态
算法
kisshyshy2 天前
🍦 雪糕、食堂、火车厢:三幅漫画吃透栈、队列与链表
javascript·算法
众少成多积小致巨2 天前
JNI (Java Native Interface) 技术手册中文参考指南
android·java·c++
猿人谷3 天前
不只是 CPU 阈值:STAR 如何用 GAT + Transformer 做容器级自动扩缩容?
人工智能·算法
复杂网络3 天前
Stable Diffusion 视觉大模型微调技术深度调研
算法