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93. 复原 IP 地址(带注释)——回溯+剪枝

guanguans 2022-01-15 阅读 38
class Solution {
public:
    vector<string> vec_str;
    vector<string> restoreIpAddresses(string s) {
        string temp = "";
        backtrack(s, 0, 0, temp);
        return vec_str;
    }
    void backtrack(string& s, int i, int n, string& temp){
        if(n == 4 || i == s.size()){
            if(n == 4 && i == s.size()) //出现4个点并且全部数字已使用,符合情况,添加
                vec_str.push_back(temp.substr(0, temp.size() - 1));
            return; //否则舍弃
        }
        for(int j = 0; j < 3; j++){
            if(j > 0 && s[i] == '0') return;    //避免"1.011.255.245"这种无效地址
            if(j == 2 && s.substr(i, 3) > "255") return;    //避免"1.111.262.45"这种无效地址
            if(i + j + 1 > s.size()) return;    //避免下标越界
            temp += s.substr(i, j + 1); //截取并添加
            temp.push_back('.');    //添加.
            //回溯
            backtrack(s, i + j + 1, n + 1, temp);
            temp = temp.substr(0, temp.size() - j - 2);
        }
    }
};

Accepted
145/145 cases passed (0 ms)
Your runtime beats 100 % of cpp submissions
Your memory usage beats 98.61 % of cpp submissions (6.2 MB)

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