Leetcode1283. 使结果不超过阈值的最小除数

二分查找，左边界为1，有边界为数组最大值

func smallestDivisor(nums []int, threshold int) int {
    left, right := 1, getMax(nums)
    for left <= right {
        mid := left + (right - left) / 2
        temp := countSum(nums, mid)
        if temp == threshold { // 如果相等，去左侧找找，看有没有更小的
            right = mid - 1
        } else if temp < threshold { // 如果太小，说明除数大了，去左侧找
            right = mid - 1
        } else { // 如果太大，说明除数太小了，去右侧找
            left = mid + 1
        }
    }
    return left
}

func countSum(nums []int, divisor int) int {
    var sum int
    for i := 0; i < len(nums); i++ {
        if nums[i] % divisor == 0 {
            sum += nums[i] / divisor
        } else {
            sum = sum + nums[i] / divisor + 1
        }
    }
    return sum
}

func getMax(nums []int) int {
    max := nums[0]
    for _, num := range nums {
        if num > max {
            max = num
        }
    }
    return max
}