Palindrome Partitioning

最新推荐文章于 2021-08-07 16:37:56 发布

原创最新推荐文章于 2021-08-07 16:37:56 发布 · 483 阅读

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本文探讨了如何利用动态规划算法解决字符串分割成回文子串的最小切割问题，通过实例演示了从简单到复杂的求解过程，包括初始化、状态转移方程的应用以及最终结果的获取。

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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.

这道题用DP来做，用一个一维数组表示从第i个元素到数组尾部的最小刀数，用一个二维数组记录从第i个元素到j个元素是否回文

class Solution {
public:
	int minCut(string s) {
		int length = s.size();
		int *cuts = new int[length + 1];
		for (int i = length; i >= 0; i--){
			cuts[i] = length - i;
		}
		bool **matrix = new bool*[length];
		for (int i = 0; i < length; i++){
			matrix[i] = new bool[length];
			memset(matrix[i], false, sizeof(bool)*length);
		}
		for (int i = length - 1; i >= 0; i--){
			for (int j = i; j < length; j++){
				if (s[i] == s[j] && (j - i < 2 || matrix[i + 1][j - 1])){
					matrix[i][j] = true;
					cuts[i] = min(cuts[i], cuts[j + 1] + 1);
				}
			}
		}
		return cuts[0] - 1;
	}
};