文章目录
一、题目
You are given n balloons, indexed from 0 to n - 1. Each balloon is painted with a number on it represented by an array nums. You are asked to burst all the balloons.
If you burst the ith balloon, you will get numsi - 1 * numsi * numsi + 1 coins. If i - 1 or i + 1 goes out of bounds of the array, then treat it as if there is a balloon with a 1 painted on it.
Return the maximum coins you can collect by bursting the balloons wisely.
Example 1:
Input: nums = 3,1,5,8
Output: 167
Explanation:
nums = 3,1,5,8 --> 3,5,8 --> 3,8 --> 8 --> \[\]
coins = 31 5 + 35 8 + 13 8 + 18 1 = 167
Example 2:
Input: nums = 1,5
Output: 10
Constraints:
n == nums.length
1 <= n <= 300
0 <= numsi <= 100
二、题解
cpp
class Solution {
public:
int maxCoins(vector<int>& nums) {
int n = nums.size();
vector<int> arr(n+2,1);
for(int i = 1;i <= n;i++){
arr[i] = nums[i-1];
}
vector<vector<int>> dp(n+2,vector<int>(n+2));
for(int i = 1;i <= n;i++){
dp[i][i] = arr[i-1] * arr[i] * arr[i+1];
}
for (int l = n, ans; l >= 1; l--) {
for (int r = l + 1; r <= n; r++) {
ans = max(arr[l - 1] * arr[l] * arr[r + 1] + dp[l + 1][r],
arr[l - 1] * arr[r] * arr[r + 1] + dp[l][r - 1]);
for (int k = l + 1; k < r; k++) {
ans = max(ans, arr[l - 1] * arr[k] * arr[r + 1] + dp[l][k - 1] + dp[k + 1][r]);
}
dp[l][r] = ans;
}
}
return dp[1][n];
}
};