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)
dfs虽然早就熟悉,但驾驭起来还是非常欠火候,而且碰到问题除非非的用dfs才会考虑,看了博主http://liaoxl.github.io/blog/20131124/leetcode-ripa/的文章,觉得dfs很强大,理解了思路,就自己实现下,仔细体会下dfs,还真的有那么点味道,这个题目不难,ip在0.0.0.0 - 255.255.255.255都是合法的,这里的程序就不做解释了,建议若看的不是很明白就自己copy到本机环境中,去单步下,很有意思,理解也更深刻,这个题本身就是一个题,但是dfs的思想还是蛮有趣。
class Solution {public: bool isValid(string sip){ if(sip.length() > 3 || sip.length() > 1 && sip[0] == '0') return false; int sum = 0; for(int i = 0; i < sip.length(); ++i){ sum *= 10; sum += sip[i] - '0'; } return (sum >= 0 && sum <= 255); } //ndot:ip地址四段三个“.",ndot表示已经处理的段数 //start:给定ip string 串的下标 void dfs(string s, int ndot, int start,string &cur_re, vector<string> &re){ if(ndot > 3 || start >= s.length()) return; if(ndot == 3 && isValid(s.substr(start))){ cur_re.append(s.substr(start)); re.push_back(cur_re); return; } for(int i = start; i < s.length() && i - start < 3; ++i){ // i - start < 3 : 4位以上可定就不合法了,不用继续了。 string sub = s.substr(start, i - start + 1); if(isValid(sub)){ int len = cur_re.length(); cur_re.append(sub + "."); dfs(s, ndot + 1, i + 1, cur_re, re); cur_re.resize(len); } } } vector<string> restoreIpAddresses(string s) { vector<string> re; string cur_re; //再少不能少于4个吧,再多也别多于12位 if(s.length() < 4 || s.length() > 12) return re; dfs(s, 0, 0, cur_re, re); return re; }};
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