LeetCode93. Restore IP Addresses

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 “[email protected]” 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

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;
    }
};