LeetCode 887 - 鸡蛋掉落

题目描述

887. 鸡蛋掉落

解法一：二分搜索（Python）

详细参考 labuladong

关键在于理解下面这张图，就理解为什么可以用到二分搜索了

class Solution:
    def superEggDrop(self, k: int, n: int) -> int:
        memo = dict()
        def dp(k, n):
            if k == 1: return n
            if n == 0: return 0
            if (k, n) in memo:
                return memo[(k, n)]
            res = float('INF')
            lo, hi = 1, n
            while lo <= hi:
                mid = (lo + hi) // 2
                broken = dp(k-1, mid-1) # 碎了
                not_broken = dp(k, n-mid) # 没碎
                if broken > not_broken:
                    hi = mid - 1
                    res = min(res, broken + 1)
                else:
                    lo = mid + 1
                    res = min(res, not_broken + 1)
            memo[(k, n)] = res
            return res

        return dp(k, n)

解法二：动态规划二（Python）

详细参考 labuladong

$dp[k][m] = n$（当前有 k 个鸡蛋，可以尝试扔 m 次鸡蛋，这个状态下，最坏情况下最多能确切测试一栋 n 层的楼）

class Solution:
    def superEggDrop(self, K: int, N: int) -> int:
        dp = [[0 for _ in range(N + 1)] for _ in range(K+1)]
        m = 0
        while dp[K][m] < N:
            m += 1
            for k in range(1, K + 1):
                dp[k][m] = dp[k][m - 1] + dp[k - 1][m - 1] + 1
        return m

解法三：滚动数组优化（Python）

class Solution:
    def superEggDrop(self, K: int, N: int) -> int:
        dp = [0] * (K + 1)
        m = 0
        while dp[K] < N:
            m += 1
            for k in range(K, 0, -1):
                dp[k] = dp[k - 1] + dp[k] + 1
        return m