Leetcode 799 champagne tower-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_43796689/article/details/109294473

本文介绍了LeetCode上的799题——香槟塔问题。当香槟倒入顶部玻璃杯后，溢出的香槟会平均分配到相邻的杯子。通过动态规划方法解决此问题，并给出代码实现。例如，倒入1杯香槟时，所有杯子都是空的；倒入2杯后，第二行的两个杯子各装满一半；倒入100000009杯后，第33行第17个杯子会装满。

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Leetcode 799 champagne tower

题目
思路
代码

题目

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 (250ml) of champagne.

Then, some champagne is poured in 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.
在这里插入图片描述
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:
$\le {\rm poured} \le 10^9$
$\le {\rm query\_glass} \le {\rm query\_row} \lt 100$

思路

考虑将上一个杯子中大于1的香槟平分到下面的两个杯子中（for循环中的内容），如果不足1了下面两个杯子即为0。最后返回对应位置的即可。要注意一个容易犯错的问题，一开始我想要如果酒很少层数很大的时候，后面一定都是0所以没有必要计算可以直接返回，但是这个判断条件是dp中的所有数字都为0，而不是poured的值大于上面几层的杯子数。例如poured = 6，query_row = 3，query_glass = 1，6 = 1+2+3，所以前面的三层杯子是够用的，但是答案不是0，因为从到二层开始每个杯子里的香槟量是不相等的，中间的先满了就会往下流但是边上的可能还有空余，需要特别考虑一下。

代码

class Solution {
    public double champagneTower(int poured, int query_row, int query_glass) {
        double[] dp = new double[query_row + 1];
        dp[0] = poured;
        for (int height = 1; height <= query_row; height++) {
            dp[height] = Math.max(0, (dp[height - 1] - 1) / 2.0);
            for (int i = height - 1; i > 0; i--) {
                dp[i] = Math.max(0, (dp[i - 1] - 1) / 2.0) + Math.max(0, (dp[i] - 1) / 2.0);
            }
            dp[0] = Math.max(0, (dp[0] - 1) / 2.0);
        }
        return Math.min(dp[query_glass], 1);
    }
}