【leetcode】 Palindrome Partitioning II-优快云博客

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46583359

本文探讨了如何使用动态规划解决最小回文切割问题，即对于给定字符串，求出将其分割为回文子串所需的最少切割次数。通过实例演示了算法实现，并解释了核心思路。

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From： https://leetcode.com/problems/palindrome-partitioning-ii/

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.

倒着

class Solution {
public:
    int minCut(string s) {
        int len = s.size();
        // i~len的最少的回文数，比cut数大1
        vector<int> d(len+1);
        // i与j是否相等
        vector<vector<bool>> m(len, vector<bool>(len, false));
        for(int i=len; i>=0; i--) d[i]=len-i;
        
        for(int i=len-1; i>=0; i--) {
            // 计算d[i]
            for(int j=i; j<len; j++) {
                //  d[i] = min(d[i], d[j+1]+1);
                //  往中间挤
                if(s[i]==s[j] && (j-i<2||m[i+1][j-1])) {
                    m[i][j] = true;
                    d[i] = min(d[i], d[j+1]+1);
                }
            }
        }
        return d[0]-1;
    }
};

顺着

public class Solution {
    public int minCut(String s) {
        int l;
		if (s == null || (l = s.length()) == 0) {
			return 0;
		}
		// 从0~i（含i）的最小回文数
		int[] palins = new int[l];
		for (int i = 0; i < l; ++i) {
			palins[i] = i + 1;
		}
		// sameAt[i][j]:i与j之间是回文
		boolean[][] sameAt = new boolean[l][l];
		for (int i = 0; i < l; ++i) {
			// 计算palins[i]
			for (int j = i; j >= 0; --j) {
				if (s.charAt(i) == s.charAt(j) && (i - j < 2 || sameAt[i - 1][j + 1])) {
					sameAt[i][j] = true;
					int palin = (j > 0 ? palins[j - 1] : 0) + 1;
					if (palin < palins[i]) {
						palins[i] = palin;
					}
				}
			}
		}
		return palins[l - 1] - 1;
    }
}