Burst Balloons （第八周 分治 + 动态规划）

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

算法思路

使用分治与动态规划了求解问题。
（1）假设输入是[a1,a2,a3,a4,a5,a6,……,an]，当我们选取第k个气球的时候，就将序列分成了两个部分，分别是N1 = [a1,a2,a3,….,a(k-1)] 与 N2 = [a(k+1),a(k+2),…..,an]。
（2）所以就把一个问题分解成两个子问题了。我们只需要知道N1和N2的最优解，再加上a（k）*a（1）*a（n），就可以求得问题的最终解。
（3）但N1和N2的最优解同样依赖于小的问题的最优解。所以我们使用自底向上的方法来求解。
（4）值得注意的是，要在原始的输入数列的头和尾插入一个1。
(5)因此，状态转移方程为：

dp[left][right] = max{dp[left][right] , nums[left] * nums[i] * nums[right] + dp[left] * nums[i] + dp[right]}*nums[i];

算法代码

class Solution {  public:     int maxCoins(vector<int>& nums) {         int dp[100][100];         int arr[100];        int n = nums.size();        for(int i = 1; i <= n; i++)            arr[i] = nums[i - 1];        arr[0] = 1;        arr[n + 1] = 1;        n += 2;        for(int i = 0 ; i < n; i++)            for(int j = 0 ; j < n; j++)                dp[i][j] = 0;        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++){                    dp[left][right] = max(dp[left][right], arr[left]*arr[i]*arr[right] + dp[left][i] + dp[i][right]);                }            }        return dp[0][n - 1];    }};