93.Restore IP Addresses
===

###### tags: `Medium`,`String`,`Backtracking`

[93. Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/)

### 題目描述

A **valid IP address** consists of exactly four integers separated by single dots. Each integer is between `0` and `255` (**inclusive**) and cannot have leading zeros.

* For example, `"0.1.2.201"` and `"192.168.1.1"` are **valid** IP addresses, but `"0.011.255.245"`, `"192.168.1.312"` and `"192.168@1.1"` are **invalid** IP addresses.

Given a string s containing only digits, return *all possible valid IP addresses that can be formed by inserting dots into* `s`. You are **not** allowed to reorder or remove any digits in `s`. You may return the valid IP addresses in **any** order.

### 範例

**Example 1:**

Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]

**Example 2:**

Input: s = "0000"
Output: ["0.0.0.0"]

**Example 3:**

Input: s = "101023"
Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]

**Constraints**:

* 1 <= `s.length` <= 20
* `s` consists of digits only.

### 解答
#### C++
##### 作法
1. backtracking走一遍

class Solution {
public:
    vector<string> ans;
    void BT(const string &s, string address = "", string cur = "", int i = 0) {
        if (i == s.size()) {
            if (cur.empty() && count(address.begin(), address.end(), '.') == 3) {
                ans.push_back(address);
            }
            return;
        }
        if (!cur.empty() && stoi(cur) == 0) return;
        cur.push_back(s[i]);
        if (stoi(cur) > 255) return;
        BT(s, address, cur, i + 1);
        if (address.empty()) {
            BT(s, cur, "", i + 1);
        } else {
            BT(s, address + "." + cur, "", i + 1);
        }
    }
    vector<string> restoreIpAddresses(string s) {
        BT(s);
        return ans;
    }
};

> [name=Yen-Chi Chen][time=Sat, Jan 21, 2023]

#### Python

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        ans = []
        def BT(address = [int(s[0])], i = 1):
            if i == len(s):
                if len(address) == 4:
                    ans.append('.'.join(map(str, address)))
                return
            old = address[-1]
            new = old * 10 + int(s[i])
            if old != 0 and new <= 255:
                address[-1] = new
                BT(address, i + 1)
                address[-1] = old
            if len(address) < 4:
                BT(address + [int(s[i])], i + 1)
        BT()
        return ans

> [name=Yen-Chi Chen][time=Sat, Jan 21, 2023]

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        def check0_255(s):
            x = int(s)
            return x>0 and x<=255

        def dfs(_s, pnts):
            #print('##', _s, pnts)
            if len(_s) > 3*(pnts+1) or len(_s) == 0:
                return []

            if pnts == 0:
                #print('##', pnts, _s, len(_s))
                if len(_s) > 4 or len(_s) == 0:
                    return []
                if _s == '0':
                    return [['0']]
                if _s[0] == '0':
                    return []
                if check0_255(_s):
                    return [ [_s] ]
                return []

            if _s[0] == '0':
                result = []
                for item in dfs(_s[1:], pnts-1):
                    #print(item)
                    result.append( ['0'] + item)
                return result

            
            result = []
            for digit in [1,2,3]:
                x = int( _s[0:digit] )
                #print(pnts, x)
                if check0_255(x):
                    for item in dfs(_s[digit:], pnts-1):
                        #print(pnts, _s, item)
                        result.append( [str(x)] + item)
            #print(pnts, _s, result)
            return result
        
        Q = dfs(s, 3)
        return [ ".".join(x) for x in Q]
        
            

### Reference

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