[Leetcode] 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.

 

Solution:

1　　首先，利用createTable来标记所有是palindrome的片段。

 

2　　初始化cut[]里的值

　

 

3　　结合state表和cut表的初始值来update。

　

public class Solution {
    public int minCut(String s) {
      if(s==null||s.length()==0||s.length()==1)
            return 0;
        int N=s.length();
        boolean[][] state=createTable(s);
        int[] cut=new int[N+1];
        cut[N-1]=0;
        for(int i=N-1;i>=0;--i){
            cut[i]=N-i;
            for(int j=i;j<N;++j){
                if(state[i][j]){
                    cut[i]=Math.min(cut[i], 1+cut[j+1]);
                }
            }
        }
        return cut[0]-1;
    }

    private boolean[][] createTable(String s) {
        // TODO Auto-generated method stub
        boolean[][] b=new boolean[s.length()][s.length()];
        for(int i=0;i<s.length();++i){
            b[i][i]=true;
        }
        int low=0;
        int high=0;
        for(int i=1;i<s.length();++i){
            //even
            low=i-1;
            high=i;
            while(low>=0&&high<s.length()&&s.charAt(low)==s.charAt(high)){
                b[low--][high++]=true;
            }
            //odd
            low=i-1;
            high=i+1;
            while(low>=0&&high<s.length()&&s.charAt(low)==s.charAt(high)){
                b[low--][high++]=true;
            }
        }
        return b;
    }
}

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2015-11-07

http://www.cnblogs.com/springfor/p/3891896.html