Leetcode212: 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

思路:

看了discuss是dp[i][j]为打破的气球为i~j之间。

我们可以想象：最后的剩下一个气球为i的时候，可以获得的分数为：nums[-1]*nums[i]*nums[n].

那么介于i,j之间的x，有： dp[i][j] = max(dp[i][j], dp[i][x – 1] + nums[i – 1] * nums[x] * nums[j + 1] + dp[x + 1][j]);

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));
        for(int l = 1; l <= n; l++)     //i和j之间的间隔l
        {
            for(int i = 1; i <= n-l+1; i++)
            {
                int j = i+l-1;
                for(int x = i; x <= j; x++) //遍历ij之间的每一种情况
                {
                    int temp = max(dp[i][j], dp[i][x-1]+dp[x+1][j]+nums[i-1]*nums[x]*nums[j+1]);
                    if(temp > dp[i][j])
                        dp[i][j] = temp;
                }
            }
        }
        return dp[1][n];
    }
};