作者： 負雪明燭
id： fuxuemingzhu
個人部落格： http://fuxuemingzhu.cn/

目錄

• 題目描述
• 題目大意
• 解題方法
• 參考資料
• 日期

題目地址：https://leetcode.com/problems/super-egg-drop/description/

題目描述

You are given K eggs, and you have access to a building with N floors from 1 to N.

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 0 <= F <= N such that any egg dropped at a floor higher than F will break, and any egg dropped at or below floor F will not break.

Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N).

Your goal is to know with certainty

What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation: 
Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.
 

Example 2:

Input: K = 2, N = 6
Output: 3

Example 3:

Input: K = 3, N = 14
Output: 4

Note:

1. 1 <= K <= 100
2. 1 <= N <= 10000

題目大意

有一個高度是N層的樓，有K個雞蛋。存在一個樓層F，使得比F高的樓層上扔下來的雞蛋都會碎，在F層以及以下的樓層扔下來的雞蛋都不會碎。每次移動我們可以使用一個雞蛋從第X層樓上扔下來，目標是找出這個F，問需要最小的移動次數是多少？

解題方法

這個題已經超出了我的能力範圍了，不過有個師兄的文章寫的超級好，對這個題分析了1萬多字，可以在這裡看到：【直觀演算法】Egg Puzzle 雞蛋難題。

class Solution:
    def superEggDrop(self, K, N):
        """
        :type K: int
        :type N: int
        :rtype: int
        """
        h, m = N, K
        if h < 1 and m < 1: return 0

        t = math.floor( math.log2( h ) ) + 1

        if m >= t: return t
        else:
            g = [ 1 for i in range(m + 1) ]
            g[0] = 0

            if g[m] >= h: return 1
            elif h == 1: return h
            else:
                for i in range(2, h + 1):
                    for j in range( m, 1, -1):
                        g[j] = g[j - 1] + g[j] + 1
                        if j == m and g[j] >= h:
                            return i
                    g[1] = i
                    if m == 1 and g[1] >= h:
                        return i

參考資料

日期

2018 年 11 月 7 日 —— 天冷加衣！