Leetcode 312. Burst Balloons

题目：

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by arraynums. You are asked to burst all the balloons. If the you burst ballooni you will get nums[left] * nums[i] * nums[right] coins. Hereleft and right are adjacent indices of i. After the burst, theleft and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

思路：

将大问题分割成小问题，对于[a1,a2,a3,a4,a5,a6,......,an]，将分割成两个子整体，分割点为k，则得到 N1 = [a1,a2,a3,....,a(k-1)], N2 = [a(k+1),a(k+2),.....,an]。这里分割点k的意义是踩破了第k个气球。于是把整体分成了两部分，问题在于，根据计算规则，k气球破了之后，a(k-1)和a(k+1)会变成相邻的，如果此时踩a(k-1)或者a(k+1)，则都会收到另一个子整体的影响，这样的话，两个子问题就不独立了。

所以设k为对于整体N，最后一个被破掉的气球。那么在k被弄破之前，k左边及右边并不会相互影响，于是就成功构造出子问题。

状态转移方程为：dp[left][right] = max(dp[left][right], dp[left][i] + nums[left] * nums[i] * nums[right] + dp[i][right]);

其中 left<i<right , dp[left][right]即为当前子问题：第left和第right之间位置的气球的maxcoin。

当然，要将nums的头和尾分别加上一个元素1。

具体代码如下：

代码：

C++实现

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