[LeetCode] 065: Palindrome Partitioning II

1.02^365的成长裂变

于 2017-09-10 20:50:32 发布

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本文介绍了一个算法问题，即如何计算给定字符串进行回文分割所需的最少切割次数，并提供了一种使用动态规划的方法来解决该问题。通过记录所有可能的回文子串，最终实现了高效的求解。

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[Problem]

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.

[Solution]

class Solution {
public:
	int minCut(string s) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		if(s.size() <= 1)return 0;

		// initial
		int dp[s.size()+1];
		memset(dp, -1, sizeof(dp));
		bool mark[s.size()][s.size()];
		memset(mark, 0, sizeof(mark));
		
		// dp
		for(int i = s.size()-1; i >= 0; --i){
			dp[i] = s.size();
			for(int j = i; j < s.size(); ++j){
				if(s[i] == s[j] && (j - i < 2 || mark[i+1][j-1])){
					mark[i][j] = true;
					dp[i] = min(dp[i], dp[j+1] + 1);
				}
			}
		}
		return dp[0];
	}
};