887.super-egg-drop.md

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 <= K <= 100
1 <= N <= 10000

前置知识

• 动态规划

思路

本题已经重制，重制版更清晰 《丢鸡蛋问题》重制版来袭～

这是一道典型的动态规划题目，但是又和一般的动态规划不一样。

拿题目给的例子为例，两个鸡蛋，六层楼，我们最少扔几次？

一个符合直觉的做法是，建立 dp[i][j], 代表 i 个鸡蛋，j 层楼最少扔几次，然后我们取 dp[K][n]即可。

代码大概这样的：

const dp = Array(K + 1);
dp[0] = Array(N + 1).fill(0);
for (let i = 1; i < K + 1; i++) {
  dp[i] = [0];
  for (let j = 1; j < N + 1; j++) {
    // 只有一个鸡蛋
    if (i === 1) {
      dp[i][j] = j;
      continue;
    }
    // 只有一层楼
    if (j === 1) {
      dp[i][j] = 1;
      continue;
    }

    // 每一层我们都模拟一遍
    const all = [];
    for (let k = 1; k < j + 1; k++) {
      const brokenCount = dp[i - 1][k - 1]; // 如果碎了
      const notBrokenCount = dp[i][j - k]; // 如果没碎
      all.push(Math.max(brokenCount, notBrokenCount)); // 最坏的可能
    }
    dp[i][j] = Math.min(...all) + 1; // 最坏的集合中我们取最好的情况
  }
}

return dp[K][N];

果不其然，当我提交的时候，超时了。 这个的时复杂度是很高的，可以看到，我们内层暴力的求解所有可能，然后 取最好的，这个过程非常耗时，大概是 O(N^2 * K).

然后我看了一位 leetcode网友的回答, 他的想法是dp[M][K]means that, given K eggs and M moves，what is the maximum number of floor that we can check.

我们按照他的思路重新建模：

可以看到右下角的部分根本就不需要计算，从而节省很多时间

关键点解析

• dp 建模思路要发生变化, 即 dp[M][K]means that, given K eggs and M moves，what is the maximum number of floor that we can check.

代码

/**
 * @param {number} K
 * @param {number} N
 * @return {number}
 */
var superEggDrop = function (K, N) {
  // 不选择dp[K][M]的原因是dp[M][K]可以简化操作
  const dp = Array(N + 1)
    .fill(0)
    .map((_) => Array(K + 1).fill(0));

  let m = 0;
  while (dp[m][K] < N) {
    m++;
    for (let k = 1; k <= K; ++k) dp[m][k] = dp[m - 1][k - 1] + 1 + dp[m - 1][k];
  }
  return m;
};