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)
分析：按照dfs的方式递归的计算，具体见代码注释 本文地址
1 class Solution { 2 public: 3 vector<string> restoreIpAddresses(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 vector<string> res; 7 string tmpres; 8 restoreIpAddressesRecur(s, 0, 4, tmpres, res); 9 return res; 10 } 11 //从s的start位置开始计算，ip地址还需要num个域（共4个域） 12 //tmpres表示当前计算好的部分ip 13 void restoreIpAddressesRecur(string &s, int start, int num, 14 string &tmpres, vector<string> &res) 15 { 16 int len = s.size(); 17 if(num == 0) 18 { 19 if(start == len)//已经计算好了一个ip 20 { 21 tmpres.erase(--tmpres.end());//去掉最后的“.” 22 res.push_back(tmpres); 23 tmpres.push_back('.'); 24 } 25 return; 26 } 27 //不满足最小或最大长度关系 28 if(len - start < num || len - start > num*3)return; 29 int tmplen = tmpres.size(); 30 string tmpstr; 31 for(int i = 1; i <= 3 && start + i <= len; i++) 32 {//ip地址的一个域最多包含三个数字 33 if(validRegion(s, start, start+i-1) == false)continue; 34 tmpstr = s.substr(start, i) + "."; 35 tmpres += tmpstr; 36 restoreIpAddressesRecur(s, start+i, num-1, tmpres, res); 37 tmpres.erase(tmplen, i+1); 38 } 39 } 40 //判断ip地址的一个域是否合法 41 bool validRegion(string &s, int istart, int iend) 42 { 43 if(iend > istart && s[istart] == '0')return false; 44 int res = 0; 45 for(int i = istart; i <= iend; i++) 46 res = res*10 + (s[i] - '0'); 47 if(res >= 0 && res <= 255)return true; 48 else return false; 49 } 50 };
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