Leetcode 93. Restore IP Addresses
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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)
s思路:
1. 正确的IP地址,有四段,每段数字在0-255之间,且每个数字不能有leading zero expect zero itself.
2. 列举所有情况,考虑backtracking,和普通的backtracking区别在判断连续的1位、2位、3位数是否在0-255之间,且没有leading zero. 如果4段内包含全部数字,则为所求。
3. 为加快速度,因为这个题限制很明确,如果给的数字长度小于4或大于12则返回空。同时,还可以判断每次尝试一个合理的字节后剩下的数字是否能构成余下的字符,不能则跳出!
//方法1:backtracking!class Solution {public: void helper(string&s,vector<string>&res,string cur,int n,int part,int idx){ if(part==4){ if(idx==n){ cur.pop_back(); res.push_back(cur); return; } } for(int i=1;i<=3;i++){ int val=stoi(s.substr(idx,i)); if(val>=0&&val<=255&&(n-idx-i)<=3*(3-part)&&(n-idx-i)>=(3-part)){ helper(s,res,cur+(s.substr(idx,i)+'.'),n,part+1,idx+i); } if(s[idx]=='0') break; } } vector<string> restoreIpAddresses(string s) { // vector<string> res; if(s.size()<4||s.size()>12) return res; helper(s,res,"",s.size(),0,0); return res; }};//取数据时,人工把多位string转换成数字,不用stoi,提升速度!class Solution {public: void helper(string&s,vector<string>&res,string cur,int n,int part,int idx){ if(part==4){ if(idx==n){ cur.pop_back(); res.push_back(cur); return; } } int val=0; for(int i=idx;i<min(3+idx,n);i++){ val=val*10+s[i]-'0'; if(val>=0&&val<=255&&(n-i-1)<=3*(3-part)&&(n-i-1)>=(3-part)){ helper(s,res,cur+(s.substr(idx,i-idx+1)+'.'),n,part+1,i+1); } if(s[idx]=='0') break; } } vector<string> restoreIpAddresses(string s) { // vector<string> res; if(s.size()<4||s.size()>12) return res; helper(s,res,"",s.size(),0,0); return res; }};
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