Question

在添加元素之前，我记录了每个元素。但是结果看起来好像覆盖了某些元素。

我不知道什么时候覆盖。

package main

import "fmt"

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

    fmt.Println(allPathsSourceTarget(graph))
}

func allPathsSourceTarget(graph [][]int) [][]int {
    n := len(graph) - 1
    result := make([][]int, 0, 200)

    var pathRecord func(target, path []int)
    pathRecord = func(target, path []int) {
        if (len(target) == 0) && (path[len(path)-1] == n) {
            fmt.Println("insert into", path) // should end with 7
            result = append(result, path)
        }

        for _, v := range target {
            pathRecord(graph[v], append(path, v))
        }
    }

    for _, v := range graph[0] {
        pathRecord(graph[v], []int{0, v})
    }

    return result
}

结果中的每个元素都应以7结尾。

Answer 1

您的问题所在：

pathRecord(graph[v], append(path, v))

Go是如此“智能”，因此他试图重用相同的分片分配的内存，而您实际上更改了已经添加到结果中的路径。）：

尝试以下方法：

newPath = make([]int, len(path))
copy(newPath, path)
pathRecord(graph[v], append(newPath, v))

Answer 2

这对我有用。我认为这很奇怪，带有追加和切片

我假定内存支持“此追加”在切片充当指针的情况下传递到递归函数中

然后下次使用相同的内存，因此它会被覆盖

因此，您需要在每次递归时复制一份，以阻止其覆盖

pathRecord = func(target, path []int) {
    if (len(target) == 0) && (path[len(path)-1] == n) {
        var c []int = make([]int, len(path))
        copy(c, path)
        //fmt.Println("insert into", payload) // should end with 7
        result = append(result, c)
    }

    for _, v := range target {
        pathRecord(graph[v], append(path, v)) //this append
    }
}

此代码何时覆盖slice元素？

2 个答案: