[LeetCode] 132. Palindrome Partitioning II Java

题目:

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.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

题意及分析:一个字符串能分割成很多回文子串,求最小分割次数。用一个数组保存当当前点的最小可能分割方法。

代码:

class Solution {
    public int minCut(String s) {
        boolean[][] isPalindrome =  new boolean[s.length()][s.length()];        //boolean[i][j]记录i到j是否为回文
        int[] cut = new int[s.length()];        //记录到当前点最小的分割次数
        for(int i=0;i<s.length();i++){
            int min = i;
            for(int j = 0;j<=i;j++){
                if(s.charAt(i)==s.charAt(j) && (i-j<=2||isPalindrome[j+1][i-1])){
                    isPalindrome[j][i] = true;
                    min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
                }
            }
            cut[i] = min;
        }

        return cut[s.length() - 1];
    }
}
原文地址:https://www.cnblogs.com/271934Liao/p/7681372.html