快速连续最大正数 [英] swift maximum consecutive positive numbers

问题描述

如何使用闭包来计算最大连续正数？

How to count maximum consecutive positive numbers using closures?

var numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]
//in this case it should be 3

print(numbers.reduce(0, { $1 > 0 ? $0 + 1 : $0 } ))//this counts total positive numbers

推荐答案

更新：更简单的解决方案：将数组拆分为
个正元素的切片，并确定最大切片长度：

Update: Simpler solution: Split the array into slices of positive elements, and determine the maximal slice length:

let  numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]
let maxConsecutive = numbers.split(whereSeparator: { $0 <= 0 }).map { $0.count }.max()!
print(maxConsecutive) // 3

---

旧答案：）使用快速累计金额：

let  numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]

let maxConsecutive = numbers.map({
    () -> (Int) -> Int in var c = 0; return { c = $0 > 0 ? c + 1 : 0; return c }
}()).max()!

此处 map（）映射每个数组元素到元素位置为止的连续正
数的计数，在这种情况下

Here map() maps each array element to the count of consecutive positive numbers up to the elements position, in this case

[1, 2, 3, 0, 0, 1, 2, 0, 0, 0, 1]

已创建转换作为立即评估
关闭以捕获变量 c ，该变量保存当前
个连续的正数。转换将递增或重置 c ，
并返回更新后的值。

The transformation is created as an "immediately evaluated closure" to capture a variable c which holds the current number of consecutive positive numbers. The transformation increments or resets c, and returns the updated value.

如果数组可能是大然后将其更改为

If the array is possibly large then change it to

let maxConsecutive = numbers.lazy.map( ... ).max()!

，以便确定最大运行长度而无需创建
中间数组。

so that the maximum run length is determined without creating an intermediate array.

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