LeetCode 132 Palindrome Partitioning II

最新推荐文章于 2021-12-18 21:08:14 发布

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本文介绍了一种使用动态规划解决LeetCode第132题“Palindrome Partitioning II”的高效方法，该方法在Java提交中击败了98.62%的提交。文章详细解释了如何通过最少的切割次数将字符串分割成回文子串。

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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.

使用动态规划 Runtime: 11 ms beats 98.62% of java submissions.

	public int minCut(String str) {
		if (str == null || str.equals("")) return 0;
		char[] chs = str.toCharArray();
		int n = chs.length;
		int[] cut = new int[n + 1];
		for (int i = 0; i <= n; i++) cut[i] = i - 1;
		for (int i = 0; i < n; i++) {
			for (int j = 0; i - j >= 0 && i + j < n && chs[i - j] == chs[i + j]; j++) {
				cut[i + j + 1] = Math.min(cut[i + j + 1], 1 + cut[i - j]);//以i为中心，奇数个字符
			}
			for (int j = 1; i - j + 1 >= 0 && i + j < n && chs[i - j + 1] == chs[i + j]; j++) {
				cut[i + j + 1] = Math.min(cut[i + j + 1], 1 + cut[i - j + 1]);//以i往前，偶数个字符
			}
		}
		return cut[n];
	}

http://www.guoting.org/leetcode/leetcode-132-palindrome-partitioning-ii/