312. Burst Balloons Hard

来源：互联网发布：乐乎公寓ifeng 编辑：程序博客网时间：2024/06/03 21:37

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

思路：本题为最优解问题，可以考虑使用动态规划。观察可知，每一次爆的气球可以把原来的数组分为两部分，一部分为所爆气球左侧的，一部分为右侧的，这样的左右两侧将互不干扰。这样可以得到问题的子问题，找到每一部分的最大值，由此得到动态规划：设定一个result[j][k],表示从气球j到k的最终计算结果的最大值，最后result[0][nums.size() + 1]表示最终结果。(由note 1,将原数组的左侧和右侧均增加一个数字为1的气球)。

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

0 0