[LeetCode] 799. Champagne Tower

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row.  Each glass holds one cup of champagne.

Then, some champagne is poured into the first glass at the top.  When the topmost glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it.  When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on.  (A glass at the bottom row has its excess champagne fall on the floor.)

For example, after one cup of champagne is poured, the top most glass is full.  After two cups of champagne are poured, the two glasses on the second row are half full.  After three cups of champagne are poured, those two cups become full - there are 3 full glasses total now.  After four cups of champagne are poured, the third row has the middle glass half full, and the two outside glasses are a quarter full, as pictured below.

Now after pouring some non-negative integer cups of champagne, return how full the jth glass in the ith row is (both i and j are 0-indexed.) 

Example 1:

Input: poured = 1, query_row = 1, query_glass = 1
Output: 0.00000
Explanation: We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.

Example 2:

Input: poured = 2, query_row = 1, query_glass = 1
Output: 0.50000
Explanation: We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.

Example 3:

Input: poured = 100000009, query_row = 33, query_glass = 17
Output: 1.00000

Constraints:

• 0 <= poured <= 109
• 0 <= query_glass <= query_row < 100

香槟塔。

我们把玻璃杯摆成金字塔的形状，其中第一层有1个玻璃杯，第二层有2个，依次类推到第100层，每个玻璃杯(250ml)将盛有香槟。

从顶层的第一个玻璃杯开始倾倒一些香槟，当顶层的杯子满了，任何溢出的香槟都会立刻等流量的流向左右两侧的玻璃杯。当左右两边的杯子也满了，就会等流量的流向它们左右两边的杯子，依次类推。（当最底层的玻璃杯满了，香槟会流到地板上）

例如，在倾倒一杯香槟后，最顶层的玻璃杯满了。倾倒了两杯香槟后，第二层的两个玻璃杯各自盛放一半的香槟。在倒三杯香槟后，第二层的香槟满了 - 此时总共有三个满的玻璃杯。在倒第四杯后，第三层中间的玻璃杯盛放了一半的香槟，他两边的玻璃杯各自盛放了四分之一的香槟，如下图所示。现在当倾倒了非负整数杯香槟后，返回第 i 行 j 个玻璃杯所盛放的香槟占玻璃杯容积的比例（i 和 j都从0开始）。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/champagne-tower
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思路是模拟，按照题意把整个金字塔里面的酒杯里面的酒量算出来即可。题意说的很清楚了，每当当前的杯子被倒满之后，多余的酒会均分到当前这个杯子下层的左右两个杯子。这个题的题设跟帕斯卡三角很像，做的时候需要把整个金字塔变成直角三角形才方便计算。假设酒的总量是能装满当前层的杯子的，那么当前的杯子装满之后，剩下的酒量 = poured - 1，再均分给两个杯子就再除以2。这样一直计算直到到达要求的那一个杯子为止。

时间O(n^2) - n是杯子的层数

空间O(n^2) - 二维的空间存储记录

Java实现

class Solution {
    public double champagneTower(int poured, int query_row, int query_glass) {
        double[][] res = new double[101][101];
        res[0][0] = (double) poured;
        for (int r = 0; r <= query_row; r++) {
            for (int c = 0; c <= r; c++) {
                double q = (res[r][c] - 1.0) / 2.0;
                if (q > 0) {
                    res[r + 1][c] += q;
                    res[r + 1][c + 1] += q;
                }
            }
        }
        return Math.min(1, res[query_row][query_glass]);
    }
}

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