快速最大连续正数_app

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{ > 0 ?

let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3

+ 1 : [+++] } ))//this counts total positive numbers

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{ > 0 ?

let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3

+ 1 : } ))//this counts total positive numbers

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{ > 0 ?

let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3

+ 1 : } ))//this counts total positive numbers

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{ > 0 ?

let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3

+ 1 : } ))//this counts total positive numbers

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{ > 0 ?

let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3

+ 1 : } ))//this counts total positive numbers

概述如何使用闭包计算最大连续正数？ var numbers = [1,3,4,-1,-2,5,2,-2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0, { $1 > 0 ? $0 + 1 : $0 } ))//this counts total positive numbers 更新：更简单的解决方案：将数组拆分为多个切片如何使用闭包计算最大连续正数？

var numbers = [1,3,4,-1,-2,5,2,-3,-4,5]//in this case it should be 3print(numbers.reduce(0,{  > 0 ? let  numbers = [1,5]let maxConsecutive = numbers.split(whereSeparator: { let  numbers = [1,5]let maxConsecutive = numbers.map({    () -> (Int) -> Int in var c = 0; return { c = [1,1,1]
 > 0 ? c + 1 : 0; return c }}()).max()!
 <= 0 }).map { let maxConsecutive = numbers.lazy.map( ... ).max()!
.count }.max()!print(maxConsecutive) // 3
 + 1 :  } ))//this counts total positive numbers

解决方法更新：更简单的解决方案：将数组拆分为多个切片
正元素,并确定最大切片长度：

老答案:)使用Swift running sum的想法：

这里map()将每个数组元素映射到连续正数的计数
在这种情况下,数字到元素位置

转换创建为“立即评估
闭包“捕获一个保存当前数字的变量c
连续的正数.转换增加或重置c,
并返回更新的值.

如果数组可能很大,则将其更改为

这样就可以确定最大运行长度而无需创建中间数组.

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