Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
   coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

一刷
題解：

我們用數組dp[lo][hi]表示從i到j的子串能夠構成的最大解
i表示子串長度，k表示在字串內the last one to burst, k in [lo, hi]，則有表達式：
dp[lo][hi] = ballons[lo-1]ballons[k]ballons[hi+1]+ dp[lo][k-1] + dp[k+1][hi])
并保存k從lo遍歷到hi時候的最大值。

public class Solution {
    public int maxCoins(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int len = nums.length + 2;
        int[] ballons = new int[len];
        ballons[0] = ballons[len-1] = 1;//boundary
        for(int i=0; i<nums.length; i++){
            ballons[i+1] = nums[i];
        }
        
        int[][] dp = new int[len][len];
        for(int i=1; i<len-1; i++){//the number of ballon to blust
            for(int lo = 1; lo<= len - i - 1; lo++){//hi - lo = i
                int hi = lo - 1 + i;
                for(int k = lo; k<=hi; k++)//lo<k<=hi, k is the last one to blust
                    dp[lo][hi] = Math.max(dp[lo][hi], ballons[lo-1]*ballons[k]*ballons[hi+1]+ dp[lo][k-1] + dp[k+1][hi]);
            }
        }
        
        return dp[1][len-2];//the answer locates in the range of [1, len-2]
    }
}