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:

• You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
• 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

/* 动态规划：
 * dp[i][j] 为打爆[i,j]内气球获得的最大奖励， 则
 * [i,j] 区间划分为： [i,k-1] k [k+1, j]
 * dp[i][j] = max(dp[i][j], nums[i-1]*nums[k]*nums[j+1]+dp[i][k-1]+dp[k+1][j])  i<=k<=j
 *                          这是因为dp[i][k-1] 表示把[i,k-1] 区间内气球全部打爆， 和k左侧相邻的气球就是i-1
 * */
class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n=nums.size();
        vector<vector<int>> dp(n+2, vector<int>(n+2, 0)); // 注意两侧是填入的哨兵
        nums.insert(nums.begin(), 1);
        nums.push_back(1);
        // nums[0]=nums[n+1]=1;  nums[1:n] 之前才是原本的数组
        // 外循环对区间长度(包含元素个数)遍历， i j 为区间两端 len= j-i+1, imin=0, jmax=n+1
        for(int len=1;len<=n;len++){
            for(int i=1;i<=n-len+1;i++){// j=len+i-1<=n  i<=n-len+1
                int j=len+i-1;
                for(int k=i;k<=j;k++){
                    dp[i][j] = max(dp[i][j], nums[i-1]*nums[k]*nums[j+1]+dp[i][k-1]+dp[k+1][j]);
                }
            }
        }
        return dp[1][n];

    }
};