[LeetCode] 131. 分割回文串

题目链接 : https://leetcode-cn.com/problems/palindrome-partitioning/

题目描述:

给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。

返回 s 所有可能的分割方案。

示例:

输入: "aab"
输出:
[
  ["aa","b"],
  ["a","a","b"]
]

思路:

思路一: 回溯算法

思路二: 动态规划 + DFS

大家可以先看 5. 最长回文子串(题解链接)的动态规划方法, 我们用dp[j][i]字符串从位置j到位置i(闭区间)是否为回文子串.

再用DFS把所有可能找到!

相关题型:132. 分割回文串 II

代码:

思路一:

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        res = []
        
        def helper(s, tmp):
            if not s:
                res.append(tmp)
            for i in range(1, len(s) + 1):
                if s[:i] == s[:i][::-1]:
                    helper(s[i:], tmp + [s[:i]])
        helper(s, [])
        return res

java

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        backtrack(res, s,  new ArrayList<String>());
        return res;

    }

    private void backtrack(List<List<String>> res, String s, ArrayList<String> tmp) {
        if (s == null || s.length() == 0) res.add(new ArrayList<>(tmp));
        for (int i = 1; i <= s.length(); i++) {
            if (isPalidrome(s.substring(0, i))) {
                // System.out.println(s.substring(0, i));
                tmp.add(s.substring(0, i));
                backtrack(res, s.substring(i, s.length()), tmp);
                tmp.remove(tmp.size() - 1);
            }
        }
    }

    private  boolean isPalidrome(String sb) {
        int left = 0;
        int right = sb.length() - 1;
        while (left < right) {
            if (sb.charAt(left) != sb.charAt(right)) return false;
            left++;
            right--;
        }
        return true;

    }
}

思路二:

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        n = len(s)
        dp = [[False] * n for _ in range(n)]

        for i in range(n):
            for j in range(i + 1):
                if (s[i] == s[j]) and (i - j <= 2 or dp[j + 1][i - 1]):
                    dp[j][i] = True
        #print(dp)
        res = []

        def helper(i, tmp):
            if i == n:
                res.append(tmp)
            for j in range(i, n):
                if dp[i][j]:
                    helper(j + 1, tmp + [s[i: j + 1]])

        helper(0, [])
        return res

java

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= i; j++) {
                if (s.charAt(i) == s.charAt(j) && (i - j < 2 || dp[j + 1][i - 1])) dp[j][i] = true;
            }
        }
        System.out.println(Arrays.deepToString(dp));
        dfs(res, dp, 0, n, s, new ArrayList<String>());
        return res;

    }

    private void dfs(List<List<String>> res, boolean[][] dp, int i, int n, String s, ArrayList<String> tmp) {
        if (i == n) res.add(new ArrayList<>(tmp));
        for (int j = i; j < n; j++) {
            if (dp[i][j]) {
                tmp.add(s.substring(i, j + 1));
                dfs(res, dp, j + 1, n, s, tmp);
                tmp.remove(tmp.size() - 1);
            }
        }
    }
}