Burst Balloons

http://www.lintcode.com/en/problem/burst-balloons/

package com.LintCode.BurstBalloons;

public class Solution {
    /*
     * @param nums: A list of integer
     * @return: An integer, maximum coins
     */
    public int maxCoins(int[] nums) {
        // write your code here
        int m = nums.length + 2;
        int[] arr = new int[m];
        System.arraycopy(nums, 0, arr, 1, m - 2);
        arr[0] = 1;
        arr[m - 1] = 1;
        int[][] dp = new int[m][m];
//        表示从i到j的最大收益
        for (int len = 1; len < m - 1; len++) {
            for (int left = 1; left + len - 1 < m - 1; left++) {
                int right = left + len - 1;
//                k是最后一个破的。所以它的收益就是乘以左边界左边和右边界的右边
                for (int k = left; k <= right; k++) {
                    dp[left][right] = Math.max(dp[left][right],
                            arr[left - 1] * arr[k] * arr[right + 1] + dp[left][k - 1] + dp[k + 1][right]);
                }
            }
        }
        return dp[1][m - 2];
    }
}