动态规划——Palindrome Partitioning II

 1 public int minCut(String s) {
 2         int slen = s.length();
 3         int[][]pa = new int[slen][slen];
 4         int[]dp = new int[slen];
 5         for(int i = 0;i<slen;i++)
 6             dp[i] = i;
 7         for(int i = 0;i<slen;i++)
 8             Arrays.fill(pa[i],0);
 9         for(int i = 1;i<slen;i++) {
10             for(int j = 0;j<=i;j++) {
11                 if(s.charAt(j)==s.charAt(i)&&((i-j<2)||pa[j+1][i-1]==1)) {
12                     pa[j][i] = 1;
13                     if(j!=0)dp[i] = Math.min(dp[i],dp[j-1]+1);
14                     else dp[i] = 0;
15                 }
16             }
17         }
18         return dp[slen-1];
19     }