leetcode 132 —— Palindrome Partitioning II

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最新推荐文章于 2021-12-18 21:08:14 发布 · 318 阅读

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本文介绍了一个使用二维DP和一维DP结合的方法来解决字符串的最小回文分割问题。通过动态规划预先计算所有子串是否为回文串，并进一步确定整个字符串达到最小分割所需的最少切割次数。

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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.

思路二维DP+一维DP

class Solution {
public:
	inline int min(int a, int b){
		return a < b ? a : b;
	}
	int minCut(string s) {
		int n = s.size();
		vector<vector<bool>> dp(n, vector<bool>(n, false));
		for (int i = 0; i < n; i++){
			for (int j = 0; j < n - i; j++){
				if (s[j] == s[j + i]){
					if (i <= 1)
						dp[j][j + i] = true;
					else
						dp[j][j + i] = dp[j + 1][j + i - 1];
				}
			}
		}
		vector<int> res(n+1, 0);
		res[n] = -1;
		for (int i = n - 1; i >= 0; i--){
			res[i] = n - i - 1;
			for (int j = i; j < n; j++){
				if (dp[i][j])
					res[i] = min(res[i], 1 + res[j + 1]);
			}
		}
		return res[0];
	}
	
};