Question

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

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

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;
  }
}

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;
  }
};

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
}

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;
  }
}