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/

題意：

給定n個氣球。每次你可以打破一個，打破第i個，那麼你會獲得nums[left] * nums[i] * nums[right]個積分。 （nums[-1] = nums[n] = 1）求你可以獲得的最大積分數

思路:

看了discuss是dp[i][j]爲打破的氣球爲i~j之間。

我們可以想象：最後的剩下一個氣球爲i的時候，可以獲得的分數爲：nums[-1]*nums[i]*nums[n].

那麼介於i,j之間的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<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 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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