【LeetCode 887】 Super Egg Drop

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 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

思路

動態規劃。dp[i][j]表示移動i次,j個雞蛋,能確定的樓層數。
dp[i][j] = dp[i-1][j-1] + dp[i-1][j] + 1。
初始化,dp[1][j] = 1,dp[1][0] = 0
倒敘循環可以壓縮到一重循環。

代碼

class Solution {
public:
    int superEggDrop(int K, int N) {
        vector<int> dp(K+1, 1);
        
        dp[0] = 0;
        int m = 1;
        
        while (dp[K] < N) {
            for (int i=K; i>=1; --i) {
                dp[i] = dp[i] + dp[i-1] + 1;
            }
            m++;
        }
        
        return m;
    }
};
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