[leetcode] Burst Balloons

题目：

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

分析：这是一个DP问题，关键是要如何转化为最优子结构问题求解，我们以最后被爆破的气球为界限，把数组分为左右两个子区域，然后分别在每个子区域中递归求解，可以用一个数组dp[i][j]记忆从下标i到下标j之间的最大coins，防止重复计算。

Java代码：

    public int maxCoins(int[] nums) {
        if(nums == null) return 0;
        int[] array = new int[nums.length+2];
        int i = 1;
        for(int num:nums) {
            if(num != 0)
                array[i++] = num;
        }
        array[0] = 1; array[i] = 1;
        int[][] dp = new int[nums.length+2][nums.length+2];
        return helper(array, dp, 0, i);
    }
    
    public int helper(int[] array, int[][] dp, int i, int j) {
        if(i+1 >= j) return 0;
        if(dp[i][j] > 0) return dp[i][j];
        int res = 0;
        for(int k = i+1; k < j; k++) {
            res = Math.max(res, array[i]*array[k]*array[j] + helper(array, dp, i, k) + helper(array, dp, k, j));
        }
        dp[i][j] = res;
        return res;
    }