312. Burst Balloons

Posted ruruozhenhao

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了312. Burst Balloons相关的知识,希望对你有一定的参考价值。

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

Example:

Input: [3, 1, 5, 8]
Output: 167
Explanation:
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

 

Approach #1: DP.[C++]

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

  

Analysis:

temp[i][j] = maxCoin(nums[i:j])

ans = temp[1][n]

temp[i][j] = max(temp[i][j], temp[i][k-1] + nums[i-1]*nums[k]*nums[j+1] + temp[k+1][j]);

 

以上是关于312. Burst Balloons的主要内容,如果未能解决你的问题,请参考以下文章

[leetcode-312-Burst Balloons]

[LeetCode] 312. Burst Balloons

LeetCode312. Burst Balloons

312. Burst Balloons

Leetcode 312. Burst Balloons

java 312. Burst Balloons(Recursive).java