Go中的旋转阵列 [英] Rotate Array in Go

问题描述

这是LeetCode问题: 189.旋转阵列:

给出一个数组，将数组向右旋转k步，其中k为非负的.

示例1:

输入:[1,2,3,4,5,6,7]且k = 3
输出:[5,6,7,1,2,3,4]

这是我的解决方案:

  func rotation(nums [] int，k int){k = k％len(数字)nums = append(nums [k:]，nums [0:k] ...)fmt.Println(数字)} 

这是一种简单明了的算法，但是不起作用.

我是Go的新手.我想 nums 是按值传递的，对 nums 的更改不会影响真正的 nums .

我该如何正确处理?

解决方案

在Go中，所有参数均按值传递.

Go切片在运行时由切片描述符表示:

  type slice struct {数组不安全len intcap int} 

如果您更改了函数中的任何切片描述符值，则通常通过返回更改后的切片描述符来传达更改.

---

您的 rotate 函数更改了指向基础数组的切片 num 指针的值和切片的容量，因此返回 num .

例如，我修复了 rotate 算法中的错误后，

 程序包主要导入"fmt"func rotation(nums [] int，k int)[] int {如果k ＜0 ||len(nums)== 0 {返回数字}fmt.Printf(数字％p数组％p len％d上限％d切片％v \ n"，& nums，& nums [0]，len(nums)，cap(nums)，nums)r:= len(数字)-k％len(数字)nums = append(nums [r:]，nums [:r] ...)fmt.Printf(数字％p数组％p len％d上限％d切片％v \ n"，& nums，& nums [0]，len(nums)，cap(nums)，nums)返回数字}func main(){nums:= [] int {1、2、3、4、5、6、7}fmt.Printf(数字％p数组％p len％d上限％d切片％v \ n"，& nums，& nums [0]，len(nums)，cap(nums)，nums)nums =旋转(nums，3)fmt.Printf(数字％p数组％p len％d上限％d切片％v \ n"，& nums，& nums [0]，len(nums)，cap(nums)，nums)} 

输出:

 数字0xc00000a080数组0xc00001a1c0 len 7 cap 7 slice [1 2 3 4 5 6 7]nums 0xc00000a0c0数组0xc00001a1c0 len 7 cap 7 slice [1 2 3 4 5 6 7]nums 0xc00000a0c0数组0xc00001a240 len 7 cap 8 slice [5 6 7 1 2 3 4]nums 0xc00000a080数组0xc00001a240 len 7 cap 8 slice [5 6 7 1 2 3 4] 

参考: The Go博客:Go Slices:用法和内部原理

This is a LeetCode problem: 189. Rotate Array:

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]

And here is my solution:

func rotate(nums []int, k int)  {
    k = k % len(nums)
    nums = append(nums[k:],nums[0:k]...)
    fmt.Println(nums)
}

It is a straight forward algorithm but it does not work.

I am new to Go. I suppose nums is passed by value and changes to nums won't affect the real nums.

How can I get this right?

解决方案

In Go, all arguments are passed by value.

A Go slice is represented at runtime by a slice descriptor:

type slice struct {
    array unsafe.Pointer
    len   int
    cap   int
}

If you change any of the slice descriptor values in a function then communicate the change, typically by returning the changed slice descriptor.

---

Your rotate function changes the values of the slice num pointer to the underlying array and the slice capacity, so return num.

For example, after I fixed the bugs in your rotate algorithm,

package main

import "fmt"

func rotate(nums []int, k int) []int {
    if k < 0 || len(nums) == 0 {
        return nums
    }

    fmt.Printf("nums %p array %p len %d cap %d slice %v\n", &nums, &nums[0], len(nums), cap(nums), nums)

    r := len(nums) - k%len(nums)
    nums = append(nums[r:], nums[:r]...)

    fmt.Printf("nums %p array %p len %d cap %d slice %v\n", &nums, &nums[0], len(nums), cap(nums), nums)

    return nums
}

func main() {
    nums := []int{1, 2, 3, 4, 5, 6, 7}

    fmt.Printf("nums %p array %p len %d cap %d slice %v\n", &nums, &nums[0], len(nums), cap(nums), nums)

    nums = rotate(nums, 3)

    fmt.Printf("nums %p array %p len %d cap %d slice %v\n", &nums, &nums[0], len(nums), cap(nums), nums)
}

Output:

nums 0xc00000a080 array 0xc00001a1c0 len 7 cap 7 slice [1 2 3 4 5 6 7]
nums 0xc00000a0c0 array 0xc00001a1c0 len 7 cap 7 slice [1 2 3 4 5 6 7]
nums 0xc00000a0c0 array 0xc00001a240 len 7 cap 8 slice [5 6 7 1 2 3 4]
nums 0xc00000a080 array 0xc00001a240 len 7 cap 8 slice [5 6 7 1 2 3 4]

Reference: The Go Blog: Go Slices: usage and internals

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