Description

Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.

Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won’t eat any more bananas during this hour.

Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.

Return the minimum integer K such that she can eat all the bananas within H hours.

Example 1:

Input: piles = [3,6,7,11], H = 8
Output: 4

Example 2:

Input: piles = [30,11,23,4,20], H = 5
Output: 30

Example 3:

Input: piles = [30,11,23,4,20], H = 6
Output: 23

Constraints:

  • 1 <= piles.length <= 10^4
  • piles.length <= H <= 10^9
  • 1 <= piles[i] <= 10^9

分析

题目的意思是：给定一个bananas数量的数组，现在一个人一次能吃k个，问最小的K使得吃的次数小于H。大概就是这么回事，首先这是一个二分查找，最小值就是1了，最大值肯定不会超过bananas的最大值，如果知道这个了，剩下的就是用二分法找一个满足条件的临界值了。

代码

class Solution:
    def solve(self,piles,k,H):
        cnt=0
        for pile in piles:
            cnt+=(pile+k-1)//k
        return cnt>H
        
    def minEatingSpeed(self, piles: List[int], H: int) -> int:
        l=1 
        h=max(piles)
        while(l<h):
            mid=l+(h-l)//2
            if(self.solve(piles,mid,H)):
                l=mid+1
            else:
                h=mid
        return l

参考文献

​​[LeetCode] solution​​