【DP】回文的最小分割数

题目描述

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.

动态规划总是练不好，强行递归每次又都超时，真的是烦。

思路：从头开始，第一个元素单独切割一定是回文。

dp[i]表示以第i个字符结尾分割成回文组的最小分割数。

isParli[i][j]表示从第i个字符到第j个字符能否构成回文。

isParli[i][j] = ((charAt(i) == charAt(k) ) && (i - k < 2 || isParli[i - 1][k + 1]))；

return最后一个dp[]即为结果。

public class Solution {        public int minCut(String s) {        int [] dp = new int [s.length() + 1];        boolean [][] isParli  = new boolean [s.length() + 1][s.length() + 1];        dp[0] = -1;                for(int i = 2; i <= s.length(); ++i)            dp[i] = Integer.MAX_VALUE;                for(int i = 1; i <= s.length(); ++i){            for(int k = 1; k <= i; ++k){                if((i - k < 2 || isParli[i - 1][k + 1]) && s.charAt(i - 1) == s.charAt(k - 1)){                    isParli[i][k] = true;                    dp[i] = Math.min(dp[i],dp[k - 1] + 1);                }            }        }        return dp[s.length()];    }    }