93. 复原 IP 地址
自己做
解：回溯法
class Solution {
private:vector<string> res;public:void tryIpAddresses(string &s, int index, int i, string &ip){ //index代表s的下标索引，i代表是第几段IP地址（分四段），ip即是保存的结果if((int)s.size() - index > (4 - i) * 3 ) //取不完完整的IP地址，比如最多取12位（3 * 4），但是字符串却有20位，这时候就取不完了，必然组不成IP地址return;if(ip.size() - 1 == s.size() + 3 && i == 4){ //保存结果【已经取完四段IP地址】ip.pop_back(); //弹出'.'res.push_back(ip);ip.push_back('.');return;}//尝试取一位ip.push_back(s[index]);ip.push_back('.');tryIpAddresses(s, index + 1, i + 1, ip);ip.pop_back();ip.pop_back();//尝试取两位if(index + 1 < (int)s.size() && s[index] > '0'){ip.push_back(s[index]);ip.push_back(s[index + 1]);ip.push_back('.');tryIpAddresses(s, index + 2, i + 1, ip);ip.pop_back();ip.pop_back();ip.pop_back();}//尝试取三位if(index + 2 < (int)s.size() && (s[index] == '1' || s[index] == '2' && s[index + 1] == '5' && s[index + 2] <= '5' || s[index] == '2' && s[index + 1] < '5')){ip.push_back(s[index]);ip.push_back(s[index + 1]);ip.push_back(s[index + 2]);ip.push_back('.');tryIpAddresses(s, index + 3, i + 1, ip);ip.pop_back();ip.pop_back();ip.pop_back();ip.pop_back();}}vector<string> restoreIpAddresses(string s) {string str;tryIpAddresses(s, 0, 0, str);return res;}
};
