[LeetCode] 875. Koko Eating Bananas 可可吃香蕉

2021年09月15日 阅读数:7
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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.html

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.python

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

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

Example 1:post

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

Example 2:this

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

Example 3:url

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

Note:spa

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

Python:code

class Solution(object):
    def minEatingSpeed(self, piles, H):
        """
        :type piles: List[int]
        :type H: int
        :rtype: int
        """
        tryEat = lambda s: sum(int(math.ceil(1.0 * p / s)) for p in piles)
        left, right = 1, sum(piles)
        while left <= right:
            mid = (left + right) / 2
            if tryEat(mid) <= H: right = mid - 1
            else: left = mid + 1
        return left 

Python: wohtm

class Solution():
    def eatBananas(self, piles, H):
        piles = sorted(piles)
        k = piles[-1]
        res = float('Inf')
        while k > 0:
            t = H
            for pile in piles:
                if pile <= k:
                    t -= 1
                else:
                    if pile % k == 0:
                        t -= pile / k
                    else:
                        t -= pile / k + 1
            if t == 0:
                res = min(res, k)
            k -= 1
        return res

  

 

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