131.Palindrome Partitioning
===

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

[131. Palindrome Partitioning](https://leetcode.com/problems/palindrome-partitioning/description/)

### 題目描述

Given a string `s`, partition s such that every
substring of the partition is a palindrome. Return *all possible palindrome partitioning of* `s`.

### 範例

**Example 1:**

Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]

**Example 2:**

Input: s = "a"
Output: [["a"]]

**Constraints**:

* 1 <= s.length <= 16
* `s` contains only lowercase English letters.

### 解答
#### C++

class Solution {
public:
    vector<vector<string>> ans;
    bool isPalindrome(const string& s) {
        int start = 0, end = s.size() - 1;
        while (start <= end) {
            if (s[start++] != s[end--]) return false;
        }
        return true;
    }
    void BT(const string& s, vector<string>& path, int cur = 0) {
        if (cur == s.size()) {
            ans.push_back(path);
            return;
        }
        for (int i = cur; i < s.size(); i++) {
            string substring = s.substr(cur, i - cur + 1);
            if (isPalindrome(substring)) {
                path.push_back(substring);
                BT(s, path, i + 1);
                path.pop_back();
            }
        }
    }
    vector<vector<string>> partition(string s) {
        vector<string> path;
        BT(s, path);
        return ans;
    }
};

> [name=Yen-Chi Chen][time=Sun, Jan 22, 2023]

#### Python

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        ans = []
        def isPalindrome(s):
            return s == s[::-1]
        def BT(path = [], index = 0):
            if index == len(s):
                ans.append(path)
                return
            for i in range(index, len(s)):
                sub = s[index:i+1]
                if isPalindrome(sub):
                    BT(path + [sub], i + 1)
        BT()
        return ans

> [name=Yen-Chi Chen][time=Sun, Jan 22, 2023]

### Reference

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