Title description
There are n balloons, numbered from 0 to n-1. Each balloon is marked with a number, and these numbers are stored in the array nums.
Now you are required to pop all the balloons. Poke the i-th balloon, you can get nums[i-1] * nums[i] * nums[i + 1] coins. Here i-1 and i + 1 represent the serial numbers of the two balloons adjacent to i. If i-1 or i + 1 exceeds the boundary of the array, then it is treated as a balloon with the number 1.
Find the maximum number of coins that can be obtained.
Example 1:
Input: nums = [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
Example 2:
Input: nums = [1,5]
Output: 10
Problem-solving ideas
class Solution_312 {
public int maxCoins(int[] nums) {
int n = nums.length;
//添加两侧的虚拟气球
int[] points = new int[n + 2];
//nums[-1] = nums[n] = 1,直接把这两个边界加进去，形成一个新的数组
points[0] = points[n +1] = 1;
for (int i = 1; i <= n; i++) {
points[i] = nums[i - 1];
}
//base case
int[][] dp = new int[n + 2][n + 2];
//开始状态转移
//i应该从下到上
for (int i = n; i >= 0; i--) {
//j应该从左到右
for (int j = i + 1; j < n + 2; j++) {
//最后戳破的气球是哪个
for(int k = i + 1; k < j; k++){
//择优做选择
dp[i][j] = Math.max(
dp[i][j],
dp[i][k] + dp[k][j] + points[i] * points[j] * points[k]
);
}
}
}
return dp[0][n + 1];
}
}