312. Burst Balloons
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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
输入一个数组nums,数组中的每一个数字代表一个气球,当选择戳破第i个气球的时候,就能获得 nums[i - 1] * nums[i] * nums[i + 1] 个金币,然后i - 1和i + 1就会成为相邻气球。问你能获得的最大金币数是多少(设nums[-1]和nums[n]均为1,但是是不可以戳破的)。
这道题可以利用动态规划来解决,首先我们要确定如何将大问题转化成小问题,设maxC[left][right]表示戳破(left, right)开区间内所有气球能获得的最大金币数,mid表示为在该区间内最后一个戳破的气球,那么就可以将问题转化为两个小问题:计算maxC[left][mid]和maxC[mid][right],当然mid可能是left和right之间的任何一个气球,因此得出状态转移方程为:
maxC[left][right] = max(maxC[left][mid] + maxC[mid][right] + nums[left] * nums[mid] * nums[right]), 其中 left<mid<right
代码如下(时间复杂度是,即O(n^3)):
class Solution {public:int maxCoins(vector<int>& nums) {int n = nums.size() + 2;nums.insert(nums.begin(), 1);nums.push_back(1);vector<vector<int>> maxC(n, vector<int>(n, 0));for (int k = 2; k < n; k++) { // k为步长for (int left = 0; left < n - k; left++) {int right = left + k;for (int mid = left + 1; mid < right; mid++)maxC[left][right] = max(maxC[left][right], maxC[left][mid] +nums[left] * nums[mid] * nums[right] + maxC[mid][right]);}}return maxC[0][n - 1];}};
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