Palindrome Partitioning II

描述

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.

分析

定义状态f(i,j)表示区间[i,j]之间最小的cut数,则状态转移方程为

f(i,j)=min{f(i,k)+f(k+1,j)},ikj,0ij<n f(i,j)=\min\left\{f(i,k)+f(k+1,j)\right\}, i \leq k \leq j, 0 \leq i \leq j<n

这是一个二维函数,实际写代码比较麻烦。

所以要转换成一维DP。如果每次,从i往右扫描,每找到一个回文就算一次DP的话,就可以转换为f(i)=区间[i, n-1]之间最小的cut数,n为字符串长度,则状态转移方程为

f(i)=min{f(j+1)+1},ij<n f(i)=\min\left\{f(j+1)+1\right\}, i \leq j<n

一个问题出现了,就是如何判断[i,j]是否是回文?每次都从i到j比较一遍?太浪费了,这里也是一个DP问题。

定义状态 P[i][j] = true if [i,j]为回文,那么

P[i][j] = str[i] == str[j] && P[i+1][j-1]

代码

// Palindrome Partitioning II
// 时间复杂度O(n^2),空间复杂度O(n^2)
public class Solution {
    public int minCut(String s) {
        final int n = s.length();
        int[] f = new int[n+1];
        boolean[][] p = new boolean[n][n];
        //the worst case is cutting by each char
        for (int i = 0; i = 0; i--) {
            for (int j = i; j < n; j++) {
                if (s.charAt(i) == s.charAt(j) && 
                        (j - i < 2 || p[i + 1][j - 1])) {
                    p[i][j] = true;
                    f[i] = Math.min(f[i], f[j + 1] + 1);
                }
            }
        }
        return f[0];
    }
}

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