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
解法
对第i个位置消去,递归之后的结果是arr[l-1]*arr[i]*arr[r+1]
dp[l][r] = max(dp[l][r], dp[l][i-1]+dp[i+1][j]+arr[l-1]*arr[r+1]*arr[i]
class Solution {
int dp[505][505];
int n;
vector<int> arr;
int dfs(int l, int r) {
if(l>r) return 0;
if(dp[l][r] > 0)
return dp[l][r];
for(int i=l;i<=r;i++) {
int left = l-1<0?1:arr[l-1];
int right = r+1>=n?1:arr[r+1];
dp[l][r] = max(dp[l][r], dfs(l, i-1) + dfs(i+1, r) + arr[i]*left*right);
}
return dp[l][r];
}
public:
int maxCoins(vector<int>& nums) {
arr = nums;
n = nums.size();
return dfs(0, n-1);
}
};