Leetcode-312. Burst Balloons

题目：

312. 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

class Solution {
public:
	int maxCoins(vector<int>& nums) {
		if (nums.size() == 1)return nums[0];
		if (nums.size() == 2)return (nums[0] * nums[1] + (nums[1] > nums[0] ? nums[1] : nums[0]));
		int size = nums.size();
		//前后加1
		nums.insert(nums.begin(), 1);
		nums.push_back(1);
		int size1 = size + 2,j=0;
		vector<long int>tmp(size1, 0);
		vector < vector < long int >> dp;
		for (int i = 0; i < size1; i++)dp.push_back(tmp);
		//填表
		for (int gap = 2; gap <size1; gap++){//间隔的最小距离
			for (int i = 0; i< size1; i++){
				j = (i + gap)>size1-1?size1-1:i+gap;
				long int temp = 0;
				for (int k = i + 1; k < j; k++){//最后一个捅破k
					temp = dp[i][k] + nums[i] * nums[k] * nums[j] + dp[k][j];
					if (temp>dp[i][j])dp[i][j] = temp;
				}
			}
		}
		return dp[0][size1 - 1];
	}
};

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分析：

1.利用分治法，当求dp[i][j]，我们假设k是(i,j)之间的开区间的数，其中k表示最后一个被弄破的气球是k时的情况，显然dp[i][j]=max{dp[i][i]+num[i]*num[k]*num[j]+dp[k][j]};

2. 用K表示气球k是最后一个弄破的，刚开始看到这道题的时候第一反应就是动态规划算法，但是只想到用dp[i][k]的k表示第一个被弄破的气球，这样对于后续转态很不好表示，

如果用k表示是最后一个弄破的气球，则很好地利用了二分法！！！