Palindrome Partitioning II

 1 public class Solution {
 2     public int minCut(String s) {
 3         int len = s.length();
 4         int D[] = new int [len+1];
 5         for(int i=0;i<=len;i++){
 6             D[i] = len-i-1; //this means for example: abc cut 2
 7         }
 8         boolean [][]p = new boolean[len][len];
 9         for(int i=len-1;i>=0;i--){
10             for(int j=i;j<len;j++){
11                 if(s.charAt(i)==s.charAt(j)&&(j-i<2 ||p[i+1][j-1])){
12                     D[i] = Math.min(D[i],D[j+1]+1);
13                     p[i][j] = true;
14                 }
15             }
16         }
17         return D[0];
18     }
19 }