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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:
- 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
本文主要参考博客:here。这道题让我再一次确认,在添加了某些边界条件后代码真的会变得非常简洁。
class Solution {
public:
int maxCoins(vector<int>& nums) {
int n = nums.size();
nums.insert(nums.begin(), 1);
nums.push_back(1);
vector<vector<int>> dp(n + 2, vector<int>(n + 2, 0));
for (int len = 1; len <= n; len++) {
for (int i = 1; i <= n - len + 1; i++) {
int j = i + len - 1;
for (int k = i; k <= j; k++) {
dp[i][j] = max(dp[i][j], nums[i - 1] * nums[k] * nums[j + 1] + dp[i][k - 1] + dp[k + 1][j]);
}
}
}
return dp[1][n];
}
};
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