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剑指 Offer II 073. 狒狒吃香蕉

题目描述

狒狒喜欢吃香蕉。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。

狒狒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，下一个小时才会开始吃另一堆的香蕉。  

狒狒喜欢慢慢吃，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

 

注意：本题与主站 875 题相同： https://leetcode.cn/problems/koko-eating-bananas/

解法

方法一

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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

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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;
    }
}

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class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        int left = 1, right = *max_element(piles.begin(), piles.end());
        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;
    }
};

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func minEatingSpeed(piles []int, h int) int {
    left, right := 1, slices.Max(piles)
    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
}

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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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class Solution {
    func minEatingSpeed(_ piles: [Int], _ h: Int) -> Int {
        var left = 1
        var right = piles.max() ?? 0

        while left < right {
            let mid = (left + right) / 2
            var hours = 0

            for pile in piles {
                hours += (pile + mid - 1) / mid
            }

            if hours <= h {
                right = mid
            } else {
                left = mid + 1
            }
        }

        return left
    }
}

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