leetcode: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)
t题目地址:https://oj.leetcode.com/problems/restore-ip-addresses/
解题思路:
题目提示用回溯法做,我没有那么做。。为啥呢。。当然是不会呀。。。等我一会儿学了回溯法在来做一遍吧。
我用了比较土的办法来做,定义三个变量loc1,loc2,loc3分别来记录三个点所在的位置。然后对切分的字符进行判断,如果4个字段都满足IP的要求就添加进结果。
对字段的判断要求:
1,可以在循环时控制前3个字段的长度(大于等于1小于等于3)(第四个字段由字符串减去前面3个字段得到)
2,字段不能以0开头,不能大于255
代码:
class Solution {public: vector<string> restoreIpAddresses(string s) {int len=s.length();vector<string> ret;if(len<=3||len>12) return ret;int loc1,loc2,loc3;for(loc1=1;loc1<=3&&loc1<len;loc1++){string s1=s.substr(0,loc1);if(atoi(s1.c_str())>255||s1.size()>1&&s1[0]=='0') continue;for(loc2=loc1+1;loc2<=loc1+3&&loc2<=len-2;loc2++){string s2=s.substr(loc1,loc2-loc1);if(stoi(s2.c_str())>255||s2.size()>1&&s2[0]=='0') continue;for(loc3=loc2+1;loc3<=loc2+3&&loc3<=len-1;loc3++){string s3=s.substr(loc2,loc3-loc2);if(atoi(s3.c_str())>255||s3.size()>1&&s3[0]=='0') continue;string s4=s.substr(loc3);if(s4.length()>3||atoi(s4.c_str())>255||s4.size()>1&&s4[0]=='0') continue;string tmp=s1+"."+s2+"."+s3+"."+s4;//cout<<tmp<<endl;ret.push_back(tmp);}}}return ret; }};
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