【LeetCode】887. Super Egg Drop 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • 参考资料
  • 日期

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

题目描述

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 the value of F is.

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

题目大意

有一个高度是N层的楼，有K个鸡蛋。存在一个楼层F，使得比F高的楼层上扔下来的鸡蛋都会碎，在F层以及以下的楼层扔下来的鸡蛋都不会碎。每次移动我们可以使用一个鸡蛋从第X层楼上扔下来，目标是找出这个F，问需要最小的移动次数是多少？

解题方法

这个题已经超出了我的能力范围了，不过有个师兄的文章写的超级好，对这个题分析了1万多字，可以在这里看到：【直观算法】Egg Puzzle 鸡蛋难题。

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

参考资料

日期

2018 年 11 月 7 日 —— 天冷加衣！