leetcode 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

原文地址（我的新blog）：http://www.hrwhisper.me/leetcode-burst-balloons/

题意：

给定n个气球。每次你可以打破一个，打破第i个，那么你会获得nums[left] * nums[i] * nums[right]个积分。 （nums[-1] = nums[n] = 1）求你可以获得的最大积分数

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]);

C++

class Solution {
public:
	int maxCoins(vector& nums) {
		int n = nums.size();
		nums.insert(nums.begin(), 1);
		nums.insert(nums.end(), 1);
		vector > dp(n + 2, vector(n + 2, 0));
		for (int k = 1; k <= n; k++) {
			for (int i = 1; i <= n - k + 1; i++) {
				int j = i + k - 1;
				for (int x = i; x <= j; x++) {
					int temp = dp[i][x - 1] + nums[i - 1] * nums[x] * nums[j + 1] + dp[x + 1][j];
					if (dp[i][j] < temp)
						dp[i][j] = temp;
				}
			}
		}
		return dp[1][n];
	}
};

java

public class Solution {
    public int maxCoins(int[] iNums) {
        int n = iNums.length;
        int[] nums = new int[n + 2];
        for (int i = 0; i < n; i++) nums[i + 1] = iNums[i];
        nums[0] = nums[n + 1] = 1;
        int[][] dp = new int[n + 2][n + 2];
        for (int k = 1; k <= n; k++) {
            for (int i = 1; i <= n - k + 1; i++) {
                int j = i + k - 1;
                for (int x = i; x <= j; x++) {
                    dp[i][j] = Math.max(dp[i][j], dp[i][x - 1] + nums[i - 1] * nums[x] * nums[j + 1] + dp[x + 1][j]);
                }
            }
        }
        return dp[1][n];
    }
}

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