LeetCode | Palindrome Partitioning II

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.

用基于备忘录（二维数组）的动态规划来纪录各个字符子串是否为回文字符串；

同时，使用一个一位数组纪录从某一索引位置开始到字符串结束位置的最小切分数。

代码

public class PalindromePartitioningII {
	public int minCut(String s) {
		if (s == null || s.length() <= 1) {
			return 0;
		}

		int N = s.length();

		// used to record whether s.subString(i, j+1) is palindrome
		boolean[][] isPalindrome = new boolean[N][N];

		// used to record the min cut of s.subString(i)
		int cut[] = new int[N + 1];
		cut[N] = -1;
		for (int i = N - 1; i >= 0; --i) {
			cut[i] = Integer.MAX_VALUE;
			for (int j = i; j < N; ++j) {
				if (s.charAt(i) == s.charAt(j)
						&& (j - i < 2 || isPalindrome[i + 1][j - 1])) {
					cut[i] = Math.min(1 + cut[j + 1], cut[i]);
					isPalindrome[i][j] = true;
				}
			}
		}

		return cut[0];
	}
}

LeetCode | Palindrome Partitioning II

原文：http://blog.csdn.net/perfect8886/article/details/19356339