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 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

题目出处：https://leetcode.com/problems/burst-balloons/

思路：易得，第一个打破的气球获得积分nums[i-1]*nums[i]*nums[i+1]，最后一个打破的气球获得积分nums[-1]*nums[i]*nums[n]。为了方便，先在数组开头和末尾都加上元素1。在区域[left,right]内，选中某个气球x最后打破，可计算此时的积分为result[left][x-1]+result[x+1][right]+nums[x]*nums[left-1]*nums[right+1]，计算出这个区域内可以得分的最大值，再逐步扩大区域范围，计算最终结果。

举个例子，数组[3,1,5]，计算最大的积分。
解题过程如下：
数组开头和末尾加元素1，变为[1,3,1,5,1]

区域长度为1时
[1,1] 最后打破气球1，resulyt[1,1]=3
[2,2] 最后打破气球2，result[2,2]=15
[3,3] 最后打破气球3，result[3,3]=5

区域长度为2时：
[1,2] 最后打破气球1,temp1=0+result[2,2]+1*3*5=30;最后打破气球2，temp2=result[1][1]+0+1*1*5=8;所以result[1,2]=30
[2,3] 最后打破气球2,temp1=0+result[3,3]+3*1*1=8;最后打破气球3，temp2=result[2][2]+0+3*5*1=30;所以result[2,3]=30

区域长度为3时：
[1,3] 最后打破气球1，temp1=0+result[2,3]+3*1*1=33;最后打破气球2，temp2=result[1,1]+result[3,3]+1*1*1=9;最后打破气球3，temp3=result[1,2]+0+5*1*1=35;所以result[1,4]=35

例题最终答案为35

思路参考：https://leetcode.com/discuss/72216/share-some-analysis-and-explanations

代码：

class Solution {
public:
    int maxCoins(vector<int>& nums) {

    int n = nums.size();  
    //数组两边 nums[0]=1,nums[n]=1
    nums.insert(nums.begin(), 1);  
    nums.insert(nums.end(), 1);

    int result[505][505]={};
    for(int length=1;length<=n;length++)
    {
        //length长度[letf,right]里找最大的
        for(int left=1;left<=(n-length+1);left++)
        {
            int right=left+length-1;
            for(int x=left;x<=right;x++)
            {
                int temp=result[left][x-1]+result[x+1][right]+nums[x]*nums[left-1]*nums[right+1];
                /*cout<<left<<" "<<right<<" "<< temp<<endl;*/
                if(temp>result[left][right])
                {
                    result[left][right]=temp;
                }
            }
        }
    }
    return result[1][n];

    }
};