README.md

875. 爱吃香蕉的珂珂

English Version

题目描述

珂珂喜欢吃香蕉。这里有 N 堆香蕉，第 i 堆中有 piles[i] 根香蕉。警卫已经离开了，将在 H 小时后回来。

珂珂可以决定她吃香蕉的速度 K （单位：根/小时）。每个小时，她将会选择一堆香蕉，从中吃掉 K 根。如果这堆香蕉少于 K 根，她将吃掉这堆的所有香蕉，然后这一小时内不会再吃更多的香蕉。  

珂珂喜欢慢慢吃，但仍然想在警卫回来前吃掉所有的香蕉。

返回她可以在 H 小时内吃掉所有香蕉的最小速度 K（K 为整数）。

 

示例 1：

输入: piles = [3,6,7,11], H = 8
输出: 4

示例 2：

输入: piles = [30,11,23,4,20], H = 5
输出: 30

示例 3：

输入: piles = [30,11,23,4,20], H = 6
输出: 23

 

提示：

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

解法

二分查找。

Python3

class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        left, right = 1, max(piles)
        while left < right:
            mid = (left + right) >> 1
            s = sum([(pile + mid - 1) // mid for pile in piles])
            if s <= h:
                right = mid
            else:
                left = mid + 1
        return left

Java

class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int mx = 0;
        for (int pile : piles) {
            mx = Math.max(mx, pile);
        }
        int left = 1, right = mx;
        while (left < right) {
            int mid = (left + right) >>> 1;
            int s = 0;
            for (int pile : piles) {
                s += (pile + mid - 1) / mid;
            }
            if (s <= h) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int mx = 0;
        for (auto pile : piles) {
            mx = max(mx, pile);
        }
        int left = 1, right = mx;
        while (left < right) {
            int mid = left + right >> 1;
            int s = 0;
            for (auto pile : piles) {
                s += (pile + mid - 1) / mid;
            }
            if (s <= h) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
};

Go

func minEatingSpeed(piles []int, h int) int {
	mx := 0
	for _, pile := range piles {
		mx = max(mx, pile)
	}
	left, right := 1, mx
	for left < right {
		mid := (left + right) >> 1
		s := 0
		for _, pile := range piles {
			s += (pile + mid - 1) / mid
		}
		if s <= h {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C#

public class Solution {
    public int MinEatingSpeed(int[] piles, int h) {
        int left = 1, right = piles.Max();
        while (left < right)
        {
            int mid = (left + right) >> 1;
            int s = 0;
            foreach (int pile in piles)
            {
                s += (pile + mid - 1) / mid;
            }
            if (s <= h)
            {
                right = mid;
            }
            else
            {
                left = mid + 1;
            }
        }
        return left;
    }
}

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