LeetCode力扣之132. Palindrome Partitioning II

最小切割实现回文串分割

最新推荐文章于 2023-05-17 11:39:00 发布

Lee_Wei4939

最新推荐文章于 2023-05-17 11:39:00 发布

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分类专栏：算法面试

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本文介绍了一种算法，用于解决字符串的最小切割问题，确保每个子串都是回文串。通过动态规划方法，文章详细阐述了如何计算并返回所需的最小切割次数。

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.

package leetCode;

/**
 * Created by lxw, liwei4939@126.com on 2018/3/26.
 */
public class L132_PalindromePartition {
    public int minCut(String s) {

        if (s == null || s.equals("")){
            return 0;
        }
        char[] chas = s.toCharArray();
        int len = chas.length;
        int[] dp = new int[len+1];
        dp[len] = -1;
        boolean[][] p = new boolean[len][len];

        for (int i = len-1; i >= 0; i--){
            dp[i] = Integer.MAX_VALUE;
            for (int j = i; j < len; j++){
                if (chas[i] == chas[j] && (j-i < 2 || p[i+1][j-1])){
                    p[i][j] = true;
                    dp[i] = Math.min(dp[i], dp[j+1] + 1);
                }
            }
        }
        return dp[0];
    }
}