93. Restore IP Addresses(dfs)
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:IP address的规则:一共四段;每段的值不能超过255;不能以0开头,但可以在一段中只有数字0
class Solution { public: vector<string> restoreIpAddresses(string s) { dfs(s, "", 1); return ret; } void dfs(string s, string currentIP, int depth){ //depth表示第几个section if(depth == 4){ if(check(s)) ret.push_back(currentIP+s); return; } int len = s.length(); if(len < 5-depth){ //剩余string长度过短 return; } string s1, s2; //check if we can assign 3 digits in the section if(len > 3 ){ s1 = s.substr(0,3); if(check(s1)){ s2 = s.substr(3); dfs(s2,currentIP+s1+".", depth+1); } } //check if we can assign 2 digits in the section if(len > 2 ){ s1 = s.substr(0,2); if(check(s1)){ s2 = s.substr(2); dfs(s2,currentIP+s1+".", depth+1); } } //assign 1 digits in the section s2 = s.substr(1); dfs(s2,currentIP+s[0]+".", depth+1); } bool check(string section){ int len = section.length(); if(len == 0 || len > 3) return false; int value = stoi(section); if(len==3){ if(section[0]!='0' && value <= 255) return true; else return false; } else if(len==2){ if(section[0]=='0') return false; else return true; } return true; } private: vector<string> ret; };
当然也能用循环,每两个section之间的分割用一个for循环遍历分割的位置,一共是三重for循环。