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
OPEN-F3 分钟前
C++进阶教程:继承与多态
开发语言·c++
鹿角片ljp2 小时前
LeetCode 236. 二叉树的最近公共祖先|递归后序
算法
luj_17682 小时前
大律师考核应重能力与科技素养
c语言·开发语言·c++·经验分享·算法
无定义_2 小时前
Floyd——Warshall
算法
刃神太酷啦2 小时前
Linux 系统 MySQL 完整安装配置教程:从卸载 MariaDB 到优化 my.cnf----《Hello MySQL!》(1)
android·linux·c语言·c++·mysql·leetcode·mariadb
带多刺的玫瑰3 小时前
Leecode#9刷题之回文数
数据结构·算法
泯泷3 小时前
手搓JSVM第 5 篇:写第一个编译器:从 AST 生成 IR
前端·javascript·算法
泯泷3 小时前
手搓JSVM第 7 篇:控制流:if、while 与 jump
前端·javascript·算法
泯泷3 小时前
手搓JSVM第 6 篇:把 IR 编成字节码:emit 与 label fixup
前端·javascript·算法
泯泷3 小时前
手搓JSVM第 3 篇:从栈式 VM 到寄存器式 VM:为什么我们选择寄存器
前端·javascript·算法