README_EN.md

132. Palindrome Partitioning II

中文文档

Description

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

 

Example 1:

Input: s = "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Example 2:

Input: s = "a"
Output: 0

Example 3:

Input: s = "ab"
Output: 1

 

Constraints:

• 1 <= s.length <= 2000
• s consists of lowercase English letters only.

Solutions

Python3

class Solution:
    def minCut(self, s: str) -> int:
        n = len(s)
        dp1 = [[False] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i, n):
                dp1[i][j] = s[i] == s[j] and (j - i < 3 or dp1[i + 1][j - 1])
        dp2 = [0] * n
        for i in range(n):
            if not dp1[0][i]:
                dp2[i] = i
                for j in range(1, i + 1):
                    if dp1[j][i]:
                        dp2[i] = min(dp2[i], dp2[j - 1] + 1)
        return dp2[-1]

Java

class Solution {
    public int minCut(String s) {
        int n = s.length();
        boolean[][] dp1 = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                dp1[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp1[i + 1][j - 1]);
            }
        }
        int[] dp2 = new int[n];
        for (int i = 0; i < n; i++) {
            if (!dp1[0][i]) {
                dp2[i] = i;
                for (int j = 1; j <= i; j++) {
                    if (dp1[j][i]) {
                        dp2[i] = Math.min(dp2[i], dp2[j - 1] + 1);
                    }
                }
            }
        }
        return dp2[n - 1];
    }
}

Go

func minCut(s string) int {
	n := len(s)
	dp1 := make([][]bool, n)
	for i := 0; i < n; i++ {
		dp1[i] = make([]bool, n)
	}
	for i := n - 1; i >= 0; i-- {
		for j := i; j < n; j++ {
			dp1[i][j] = s[i] == s[j] && (j-i < 3 || dp1[i+1][j-1])
		}
	}
	dp2 := make([]int, n)
	for i := 0; i < n; i++ {
		if !dp1[0][i] {
			dp2[i] = i
			for j := 1; j <= i; j++ {
				if dp1[j][i] {
					dp2[i] = min(dp2[i], dp2[j-1]+1)
				}
			}
		}
	}
	return dp2[n-1]
}

func min(x, y int) int {
	if x < y {
		return x
	}
	return y
}

C++

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> dp1(n, vector<bool>(n));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                dp1[i][j] = s[i] == s[j] && (j - i < 3 || dp1[i + 1][j - 1]);
            }
        }
        vector<int> dp2(n);
        for (int i = 0; i < n; ++i) {
            if (!dp1[0][i]) {
                dp2[i] = i;
                for (int j = 1; j <= i; ++j) {
                    if (dp1[j][i]) {
                        dp2[i] = min(dp2[i], dp2[j - 1] + 1);
                    }
                }
            }
        }
        return dp2[n - 1];
    }
};

...