Question

我正在尝试生成列表的所有排列，但每次我递归列表时，我都会重置回原来的状态，那是什么？

public static void PermuteAndSolve(List<City> cities, int recursionLevel, int Length)
    {
        if (recursionLevel == Length)
        {
            foreach (City city in cities)
            {
                Console.Write(city.name);
            }
            Console.WriteLine("");
        }
        else
        {
            for (int i = recursionLevel; i <= Length; i++)
            {
                swap(cities, recursionLevel, Length);
                PermuteAndSolve(cities, recursionLevel + 1, Length);
                swap(cities, recursionLevel, Length);


            }
        }
    }

我知道交换工作正常。

Answer 1

我并没有假装准确理解这段代码的作用，但是如果没有令人费解的递归，你肯定可以实现你想要做的事情吗？这样的事情（注意：未经测试）：

public static void PermuteAndSolve(List<City> cities, int Length)
{
    int recursionLevel = 0;
    while(recursionLevel < Length)
    {
        for (int i = recursionLevel; i <= Length; i++) {
            swap(cities, recursionLevel, Length);
        }
        recursionLevel++;
    }

    foreach(City city in cities)
    {
        Console.Write(city.name);
    }
    Console.WriteLine("");
}

可能不正确，但希望你明白我的观点。

Answer 2

原来我是一个白痴，我在意外递归之后添加了额外的交换。真实代码应为：

public static void PermuteAndSolve(List<City> cities, int recursionLevel, int Length)
    {
        if (recursionLevel == Length)
        {
            foreach (City city in cities)
            {
                Console.Write(city.name);
            }
            Console.WriteLine("");
        }
        else
        {
            for (int i = recursionLevel; i <= Length; i++)
            {
                swap(cities, recursionLevel, Length);
                PermuteAndSolve(cities, recursionLevel + 1, Length);                    
            }
        }
    }

尝试置换时，在每次递归时列出重置

2 个答案: