【leetcode】1283. Find the Smallest Divisor Given a Threshold

题目如下：

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Constraints:

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

解题思路：除数越大，计算出来的和越小，因此可以使用二分查找法。

代码如下：

class Solution(object):
    def smallestDivisor(self, nums, threshold):
        """
        :type nums: List[int]
        :type threshold: int
        :rtype: int
        """
        res = 0
        low,high = 1,1000000
        while low <= high:
            mid = (low + high)/2
            count = 0
            for i in nums:
                count += (i/mid)
                if i%mid != 0:count += 1
            if count <= threshold:
                res = mid
                high = mid - 1
            else:
                low = mid + 1
        return res

原文地址：https://www.cnblogs.com/seyjs/p/12026288.html