Palindrome Partitioning II

最新推荐文章于 2021-06-11 12:04:51 发布

最新推荐文章于 2021-06-11 12:04:51 发布 · 62 阅读

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本文介绍了一个算法问题：如何计算将字符串分割成多个回文子串所需的最少切割次数，并提供了一个Java实现示例。

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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.

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