LeetCode93—Restore IP Addresses

类似路径搜索问题，IP地址每一小段的范围是0~255，对于输入的字串来说，考虑长度分别为1，2,3这三种情况。

原题

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

想象一个树。一共四层，每一层代表ip地址的小段，每一层必须满足条件”0~255”（1位数到3位数），且不能出现”01” “024”这种首字母为0的情况。
递归搜索结果，返回的条件：
1. 收集满4段，并且每段有效
2. 输入串s的字符全部用完
对于第2个条件，我之前的想法是，当收集满4段之后，把这4段长度加起来与s.size()比较，后来网上看到一种更好的方法：让s.size()和index（递归变量）比较，这样效率更高，不用每次循环计算长度，均在代码中注释。

代码

class Solution {
private:bool valid(string s){if (s.size() == 2){if (s[0] == '0')//"01"情况排除return false;}if (s.size() == 3){if (s[0] == '0' || atoi(s.c_str()) > 255)//除了"001"还要排除大于255return false;}return true;}void help(string s, int len, vector<string>& tmpRes, vector<string>&result, int index){if (tmpRes.size() == 4){//当满足条件时，s里面的字符肯定用完，i==s.size()-1，则index==i+1==s.size()if (index == s.size()){string ip;for (int i = 0; i < tmpRes.size(); i++){ip += (tmpRes[i] + '.');}ip.pop_back();result.push_back(ip);}return;//下述方法每次都计算ip的长度，效率较低/*string ip;for (int i = 0; i < tmpRes.size(); i++){ip += (tmpRes[i] + '.');}ip.pop_back();if (ip.size() == len + 3){result.push_back(ip);}elsereturn;*/}string str;for (int i = index; (i < len) && (str.size()<3); i++){str.push_back(s[i]);if (valid(str)){tmpRes.push_back(str);help(s, len, tmpRes, result, i + 1);tmpRes.pop_back();}}}
public:vector<string> restoreIpAddresses(string s) {vector<string>result;vector<string>tmpRes;help(s, s.size(), tmpRes, result, 0);return result;}
};

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LeetCode93—Restore IP Addresses

LeetCode93—Restore IP Addresses

原题

分析

代码

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