Leetcode: Restore IP Addresses
2014-12-02 21:45
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Given a string containing only digits, restore it by returning all possible valid IP address combinations.
For example:
Given
return
分析:DFS+Backtracking,这道题是剪枝比较典型的例子。但相比其他类似问题,这道题有更多的限制条件,在写代码前必须对这些限制条件有清晰的了解。
(1)ip地址必须是4段。此条件可以用来减枝,因为ip的每一段最少有一个字符最多有三个字符。
(2)ip地址每一段位于范围[0,255]。
(3)ip地址的每一段不能有preceding的0,但该段只有一个0是可以的。
代码如下:
For example:
Given
"25525511135",
return
["255.255.11.135", "255.255.111.35"]. (Order does not matter)
分析:DFS+Backtracking,这道题是剪枝比较典型的例子。但相比其他类似问题,这道题有更多的限制条件,在写代码前必须对这些限制条件有清晰的了解。
(1)ip地址必须是4段。此条件可以用来减枝,因为ip的每一段最少有一个字符最多有三个字符。
(2)ip地址每一段位于范围[0,255]。
(3)ip地址的每一段不能有preceding的0,但该段只有一个0是可以的。
代码如下:
class Solution { public: vector<string> restoreIpAddresses(string s) { vector<string> result; string ip; dfs(result, ip, s, 0, 0); return result; } void dfs(vector<string> &result, string ip, string s, int start, int step){ if(start == s.length() && step == 4){ ip.pop_back();//remove trailing '.' result.push_back(ip); return; } if(s.size() - start > (4-step)*3) return; //prune if(s.size() - start < 4-step) return; //prune int num = 0; for(int i = start; i < start+3; i++){ num = num*10 + (s[i] - '0'); if(num <= 255){ ip += s[i]; dfs(result, ip+'.', s, i+1, step+1); } if(num == 0) break; // no preceding zero allowed but only one zero } } };
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