当前位置：首页 > 文章列表 > Golang > Go问答 > 为什么我的范围在切片中显示不存在的值？

为什么我的范围在切片中显示不存在的值？

来源：stackoverflow 2024-04-20 13:42:36 0浏览收藏

珍惜时间，勤奋学习！今天给大家带来《为什么我的范围在切片中显示不存在的值？》，正文内容主要涉及到等等，如果你正在学习Golang，或者是对Golang有疑问，欢迎大家关注我！后面我会持续更新相关内容的，希望都能帮到正在学习的大家！

问题内容

我试图用 go 解决 leetcode 问题。问题是子集。

这是我正在编写的完整代码以及一些调试日志：

package main

import (
    "fmt"
)

func main() {
    v := []int{9, 0, 3, 5, 7}
    fmt.println(subsets(v))
}

func subsets(nums []int) [][]int {
    result := [][]int{
        []int{}, // empty
    }

    for _, num := range nums {  
        fmt.println("==========")
        fmt.println(num)
        fmt.printf("result = %v\n", result)

        temp := [][]int{}
        for _, r := range result {
            fmt.printf("r = %v\n", r)
            temp = append(temp, append(r, num))
        }

        for _, t := range temp {
            result = append(result, t)
        }

        fmt.println("==========")       
    }

    return result
}

（我还准备了go游乐场网址）

这是上面代码的输出：

==========
9
result = [[]]
r = []
==========
==========
0
result = [[] [9]]
r = []
r = [9]
==========
==========
3
result = [[] [9] [0] [9 0]]
r = []
r = [9]
r = [0]
r = [9 0]
==========
==========
5
result = [[] [9] [0] [9 0] [3] [9 3] [0 3] [9 0 3]]
r = []
r = [9]
r = [0]
r = [9 0]
r = [3]
r = [9 3]
r = [0 3]
r = [9 0 3]
==========
==========
7
result = [[] [9] [0] [9 0] [3] [9 3] [0 3] [9 0 3] [5] [9 5] [0 5] [9 0 5] [3 5] [9 3 5] [0 3 5] [9 0 3 5]] // (a)
r = []
r = [9]
r = [0]
r = [9 0]
r = [3]
r = [9 3]
r = [0 3]
r = [9 0 3]
r = [5]
r = [9 5]
r = [0 5]
r = [9 0 5]
r = [3 5]
r = [9 3 5]
r = [0 3 5]
r = [9 0 3 7] // (b)
==========
[[] [9] [0] [9 0] [3] [9 3] [0 3] [9 0 3] [5] [9 5] [0 5] [9 0 5] [3 5] [9 3 5] [0 3 5] [9 0 3 7] [7] [9 7] [0 7] [9 0 7] [3 7] [9 3 7] [0 3 7] [9 0 3 7] [5 7] [9 5 7] [0 5 7] [9 0 5 7] [3 5 7] [9 3 5 7] [0 3 5 7] [9 0 3 7 7]]

让我们看看第五个 result。（我在那里指出了 (a) ）到目前为止， result 的最后一个元素是 [9, 0, 3, 5] ，这是预期的行为。但是，之后，当我尝试在 result 中写入该元素的调试日志时，它会更改为 [9, 0, 3, 7] ( (b) )。

你知道为什么吗？

解决方案

append 在需要时更改 r：
而不是：

temp = append(temp, append(r, num))

您可以使用：

rr := make([]int, len(r))
            copy(rr, r)
            rr = append(rr, num)
            temp = append(temp, rr)

你就可以开始了：

package main

import (
    "fmt"
)

func main() {
    v := []int{9, 0, 3, 5, 7}
    fmt.println(subsets(v))
}

func subsets(nums []int) [][]int {
    result := [][]int{
        []int{}, // empty
    }

    for _, num := range nums {
        fmt.println("==========")
        fmt.println(num)
        fmt.printf("result = %v\n", result)

        temp := [][]int{}
        for _, r := range result {
            fmt.printf("r = %v\n", r)
            // append(r, num)
            rr := make([]int, len(r))
            copy(rr, r)
            rr = append(rr, num)
            temp = append(temp, rr)
        }

        for _, t := range temp {
            result = append(result, t)
        }

        fmt.println("==========")
    }

    return result
}

只需调试您的代码，您就会看到，它发生在这里：

temp := [][]int{}
        for _, r := range result {
            fmt.println("result =", result, len(result), cap(result))
            fmt.println("r ==", r, len(r), cap(r))
            fmt.println("num =", num)
            rr := append(r, num)
            fmt.println("r ==", r, len(r), cap(r))
            fmt.println("rr ==", rr, len(rr), cap(rr))
            fmt.println("result =", result, len(result), cap(result))
            fmt.println("temp =", temp)
            temp = append(temp, rr)
            fmt.println("temp =", temp)
        }

什么时候

result = [[] [9] [0] [9 0] [3] [9 3] [0 3] [9 0 3] [5] [9 5] [0 5] [9 0 5] [3 5] [9 3 5] [0 3 5] [9 0 3 5]] 16 16
r == [9 0 3] 3 4
num = 7
r == [9 0 3] 3 4
rr == [9 0 3 7] 4 4
result = [[] [9] [0] [9 0] [3] [9 3] [0 3] [9 0 3] [5] [9 5] [0 5] [9 0 5] [3 5] [9 3 5] [0 3 5] [9 0 3 7]] 16 16

因为：