Palindrome Partitioning II

public class Solution {
public int minCut(String s) {
        int n = s.length();
        boolean[][] isPalindrome = new boolean[n][n];
        for (int i = 0; i < n; i++)
            isPalindrome[i][i] = true;
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i + 1; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    if (j - i < 2 || isPalindrome[i + 1][j - 1])
                        isPalindrome[i][j] = true;
                }
            }
        }
        int[] minCut = new int[n + 1];
        for (int i = n; i >= 0; i--)
            minCut[i] = n - 1 - i;
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (isPalindrome[i][j]) {
                    minCut[i] = Math.min(minCut[i], 1 + minCut[j + 1]);
                }
            }
        }
        return minCut[0];
    }
}