[LeetCode] 93. Restore IP Addresses

Restore IP Addresses

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

Input: “25525511135”
Output: [“255.255.11.135”, “255.255.111.35”]

解法1：暴力搜索

把字符串分为4段，每一段的长度小于4，大于0，遍历四个长度，当四个长度和为字符串的大小时，保存。

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        for(int a=1;a<4;a++){
            for(int b=1;b<4;b++){
                for(int c=1;c<4;c++){
                    for(int d=1;d<4;d++){
                        if(a+b+c+d == s.size()){
                            int n1 = stoi(s.substr(0,a));
                            int n2 = stoi(s.substr(a,b));
                            int n3 = stoi(s.substr(a+b,c));
                            int n4 = stoi(s.substr(a+b+c,d));
                            if(n1<=255 && n2<=255 && n3<=255 && n4<=255){
                                string t = to_string(n1)+"."+to_string(n2)+"."+to_string(n3)+"."+to_string(n4);
                                if(t.size() == s.size()+3)
                                    res.push_back(t);
                            }                            
                        }

                    }
                }
            }
        }
        return res;
    }
};

解法2：递归

定义一个数表示递归的段数，可以从4到0，也可以从0到4，并判断字符串是否合法。
从0->4

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        helper(s, 0, "",res);
        return res;
    }
    void helper(string s, int k, string out, vector<string>& res){
        if(k == 4){
            if(s.empty())
                res.push_back(out);            
        }
        else{
            for(int i=1;i<=3;i++){
                if(s.size() < i) break;
                int val = stoi(s.substr(0,i));
                if(val > 255 || to_string(val).size() != i)
                    continue;
                if(k==3)
                    helper(s.substr(i), k+1, out+s.substr(0,i),res);
                else
                    helper(s.substr(i), k+1, out+s.substr(0,i)+".",res);
            }            
        }
    }
};

4->0

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        helper(s, 4, "",res);
        return res;
    }
    void helper(string s, int k, string out, vector<string>& res){
        if(k == 0){
            if(s.empty())
                res.push_back(out);            
        }
        else{
            for(int i=1;i<=3;i++){
                if(s.size()>=i && isValid(s.substr(0,i))){
                    if(k==1)
                        helper(s.substr(i), k-1, out+s.substr(0,i),res);
                    else
                        helper(s.substr(i), k-1, out+s.substr(0,i)+".",res);
                }
            }            
        }

    }
    bool isValid(string s){
        if (s.empty() || s.size() > 3 || (s.size() > 1 && s[0] == '0')) return false;
        int res = stoi(s);
        return res <= 255 && res >= 0;
    }
};

参考

https://www.cnblogs.com/grandyang/p/4305572.html