Palindrome Partitioning II

Palindrome Partitioning II

问题：

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

思路：

　　动态规划

我的代码1：（思路和代码都是正确的，但是对于大数据超时啊） http://www.lintcode.com/zh-cn/problem/palindrome-partitioning-ii/ 都通过不了啊

public class Solution {
    public int minCut(String s) {
        if(s==null || s.length()==0 || isPalindrome(s))    return 0;
        int len = s.length();
        int[] record = new int[len];
        
        for(int i=0; i<len; i++)
        {
            int min = Integer.MAX_VALUE;
            for(int j=0; j<=i; j++)
            {
                String sub = s.substring(j,i+1);
                if(isPalindrome(sub))
                {
                    min = Math.min(min, (j==0?0:1+record[j-1]));
                }
            }
            record[i] = min;
        }
        return record[len-1];
        
    }
    public boolean isPalindrome(String part)
    {
        int len = part.length();
        for(int i = 0; i < len/2; i++)
        {
            char first = part.charAt(i);
            char last = part.charAt(len - 1 - i);
            if(first != last)   return false;
        }
        return true;
    }
}

 我的代码2：用另外一个数组记录已经成功的substring http://www.lintcode.com/zh-cn/problem/palindrome-partitioning-ii/ 可以通过了

public class Solution {
    public int minCut(String s) {
        if(s==null || s.length()==0 || isPalindrome(s))    return 0;
        int len = s.length();
        int[] record = new int[len];
        boolean[][] isValid = new boolean[len][len];
        
        for(int i=0; i<len; i++)
        {
            int min = Integer.MAX_VALUE;
            for(int j=0; j<=i; j++)
            {
                String sub = s.substring(j,i+1);
                if(isValid[j][i])
                {
                    min = Math.min(min, (j==0?0:1+record[j-1]));
                    continue;
                }
                if(isPalindrome(sub))
                {
                    min = Math.min(min, (j==0?0:1+record[j-1]));
                    isValid[j][i] = true;
                }
            }
            record[i] = min;
        }
        return record[len-1];
        
    }
    public boolean isPalindrome(String part)
    {
        int len = part.length();
        for(int i = 0; i < len/2; i++)
        {
            char first = part.charAt(i);
            char last = part.charAt(len - 1 - i);
            if(first != last)   return false;
        }
        return true;
    }
}