312.Burst Balloons

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 ballooni 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 1671

    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

Credits:
Special thanks to @peisi for adding this problem and creating all test cases.

像这种求极值问题，我们一般都要考虑用动态规划Dynamic Programming来做，我们维护一个二维动态数组dp，其中dp[i][j]表示打爆区间[i,j]中的所有气球能得到的最多金币。题目中说明了边界情况，当气球周围没有气球的时候，旁边的数字按1算，这样我们可以在原数组两边各填充一个1，这样方便于计算。这道题的最难点就是找递归式，如下所示：

dp[i][j] = max(dp[i][j], nums[i - 1]*nums[k]*nums[j + 1] + dp[i][k - 1] + dp[k + 1][j])                  i∈[1,n],j∈[1,n-i+1] k∈[j,j+i)

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n=nums.size();
        nums.insert(nums.begin(),1);
        nums.insert(nums.end(),1);
        vector<vector<int>>dp(n+2,vector<int>(n+2,0));
        //dp[i][j]表示i到j求得的最大coin值
        for(int len=1;len<=n;len++){//计算的气球下标范围长度
            for(int start=1;start<=n-len+1;start++){//气球下标开始的位置范围
                int end=start+len-1;//气球下标结束的位置
                for(int x=start;x<=end;x++){
                    dp[start][end]=max(dp[start][end],dp[start][x-1]+nums[start-1]*nums[x]*nums[end+1]+dp[x+1][end]);
                }
            }
        }
        return dp[1][n];
    }
};