【leetcode】Restore IP Addresses (middle)

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

 

思路：回溯法 解向量X={str0, str1, str2, str3} 分别是IP地址的四段

判断每个解向量的部分时，先求出该部分最长和最短长度（根据后面每段最少1位，最多3位），回溯求解。符合条件时就把X按照要求的格式压入ans.

注意，判断每个部分是否有效时要排除00，01，001...等0开头的情况。

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> ans;
        int len = s.length();
        if(len < 4)
            return ans;
        
        vector<string> X(4, "");
        vector<vector<string>> S(4, vector<string>());
        int k = 0;
        int maxlen = min(3, len - 3 * 1);
        int minlen = max(1, len - 3 * 3);
        for(int i = minlen; i <= maxlen; i++)
        {
            string sub = s.substr(0, i);
            if(isVaildIP(sub))
            {
                S[0].push_back(sub);
            }
        }

        while(k >= 0)
        {
            while(!S[k].empty())
            {
                X[k] = S[k].back();
                S[k].pop_back();
                if(k < 3)
                {
                    k = k + 1;
                    int startloc = 0;
                    for(int j = 0; j < k; j++)
                    {
                        startloc += X[j].length();
                    }
                    int maxlen = min(3, len - (4 - k - 1) * 1 - startloc);
                    int minlen = max(1, len - (4 - k - 1) * 3 - startloc);
                    for(int i = minlen; i <= maxlen; i++)
                    {    
                        string sub = s.substr(startloc, i);
                        if(isVaildIP(sub))
                        {
                            S[k].push_back(sub);
                        }
                    }
                }
                else
                {
                    ans.push_back(X[0]);
                    for(int i = 1; i < 4; i++)
                    {
                        ans.back().append(".");
                        ans.back().append(X[i]);
                    }
                }
            }
            k--;
        }

        return ans;
    }

    bool isVaildIP(string snum)
    {
        if(snum.size() > 1 && snum[0] == '0')
            return false;
        int num = atoi(snum.c_str());
        return (num >= 0 && num <= 255);
    }
};

int main()
{
    Solution s;
    string num =/* "25525511135"*/"010010";
    vector<string> ans = s.restoreIpAddresses(num);

    return 0;
}