131.Palindrome Partitioning

tags: Medium,String,Backtracking,DP

131. Palindrome Partitioning

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;
    }
};

Yen-Chi ChenSun, 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

Yen-Chi ChenSun, Jan 22, 2023

Reference

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