题目描述(Hard)

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.

题目链接

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

Example 1:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

算法分析

每次从i往右扫描，每找到一个回文就算一次DP，可以转换为f(i)=[i, n-1]之间的最小的cut数，n为字符串长度，则状态转移方程为。判断[i，j]为回文，每次从i到j比较太过于费时，可以定义状态P[i][j]=true,如果[i,j]为回文，那么P[i][j]=str[i]==str[j] && P[i+1][j-1]。

提交代码：

class Solution {
public:
    int minCut(string s) {
        const int n = s.size();
        int f[n + 1];
        bool p[n][n];
        
        fill_n(&p[0][0], n * n, false);
        for (int i = 0; i <= n; ++i)
            f[i] = n - i - 1;
        
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                if (s[i] == s[j] && (j - i < 2 || p[i + 1][j - 1])) {
                    p[i][j] = true;
                    f[i] = min(f[i], f[j + 1] + 1);
                }
            }
        }
        
        return f[0];
    }
};