编程练习（第十二周）

题目来源于:https://leetcode.com

312. Burst Balloons

Add to List

DescriptionHintsSubmissionsSolutions

• Total Accepted: 24690
• Total Submissions: 58469
• Difficulty: Hard
• Contributor: LeetCode

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

题解：

class Solution {public:    vector<vector<int>> d;    int maxCoins(vector<int>& nums) {        if(nums.empty()){            return 0;        }        int size = nums.size();        nums.insert(nums.begin(), 1);        nums.insert(nums.end(), 1);        d = vector<vector<int>>(size + 2, vector<int>(size + 2, 0));        for(int len = 1; len <= size; ++len){            for(int start = 1;start + len - 1 <= size; ++start){                int end = start + len - 1;                for(int k = start;k <= end; ++k){// k is the last one to be popped                    d[start][end] = max(d[start][end], d[start][k - 1] + d[k + 1][end] + nums[start - 1] * nums[k] * nums[end + 1]);                }            }        }        return d[1][size];    }};