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
YL200404265 小时前
048路径总和III
数据结构·dfs
z200509305 小时前
每日简单算法题——————跟着卡尔
算法
Irissgwe6 小时前
类与对象(三)
开发语言·c++·类和对象·友元
️是786 小时前
信息奥赛一本通—编程启蒙(3395:练68.3 车牌问题)
数据结构·c++·算法
Liangwei Lin6 小时前
LeetCode 118. 杨辉三角
算法·leetcode·职场和发展
计算机安禾6 小时前
【c++面向对象编程】第24篇:类型转换运算符:自定义隐式转换与explicit
java·c++·算法
鼠鼠我(‘-ωก̀ )好困6 小时前
leetGPU
算法
雪度娃娃6 小时前
转向现代C++——优先选用nullptr而不是0和NULL
开发语言·c++
我星期八休息7 小时前
Linux系统编程—基础IO
linux·运维·服务器·c语言·c++·人工智能·算法
池塘的蜗牛7 小时前
A Low-Complexity Method for FFT-based OFDM Sensing
算法