Burst Balloons——Difficulty：Hard

Problem :

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

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Algorithm:

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动态规划：arr[left][right]表示区间（left，right）之间能得到的最大分数值，状态转移方程把第i个气球看作是最后一个踩爆的气球，这样这个气球踩爆后是不会影响后续的状态。
状态转移方程：arr[left][right] = max(arr[left][right],nums[left]*nums[i]*nums[right] + arr[left][i] + arr[i][right]);

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Code：

class Solution {public:    int maxCoins(vector<int>& nums) {        nums.push_back(1);        nums.insert(nums.begin(),1);        int arr[nums.size()][nums.size()]={};        int n=nums.size();        for(int k=2;k<n;k++)        {            for(int left = 0;left<n-k;++left){                  int right = left + k;                  for(int i=left+1;i< right; ++i)                  {                      arr[left][right] = max(arr[left][right],nums[left]*nums[i]*nums[right] + arr[left][i] + arr[i][right]);                  }              }              }        return arr[0][nums.size()-1];    }};