下面的算法，并给出我们应该把它写在Java中。然而，当我试图理解一行行，它变得混乱，特别是部分：

The following algorithm is given and we are supposed to write it out in java. However, when I try to understand line by line, it gets confusing, especially the part:

A [k + 1：N-1] =升序排列S中的值

A[k+1:N-1] = the values in S in ascending order

要我明白了，只设置有1个号码在任何时候。我们如何能取代 A [k + 1：N-1] 的设置时，只有1号

To my understand, the set only have 1 number at anytime. How can we replace A[k+1:N-1] when the set only has 1 number?

设A是按升序排列（假设它是整数数组 0 到 N-1 序列对 INT [N] ）。

Let A be a sequence of integers 0 to N-1 in ascending order (let's assume its an array of int[N]).

next_permutation(A):
    k = N-1
    S = { }
    while k >= 0:
        if S contains a value larger than A[k]:
            v = the smallest member of S that is larger than A[k]
            remove v from S
            insert A[k] in S
            A[k] = v
            A[k+1:N-1] = the values in S in ascending order.
            return true
        else:
            insert A[k] in S
            k -= 1
    return false

推荐答案

所示的算法是类似于生成字典顺序。你可以阅读这篇文章以获取更多信息。

The algorithm shown is similar to the Generation in lexicographic order. You can read the article for further information.

@templatetypedef如何我可以值一组按升序排列替换数组中的项目任何线索？例如A [k + 1：N-1] =升序排列S中的值。我应该使用的toArray（）？

@templatetypedef Any clue on how I can replace items in an array by values in a set in ascending order? e.g. A[k+1:N-1] = the values in S in ascending order. Should I use toArray()?

这是没有必要的。你可以尽量保持S数组排序的所有的时间。要插入阵列中的一个新的号码时，都将其插入这样的，所以阵列能留下来分类的。 例如，如果你有 S = [5 7 8] 到目前为止，你要插入 6 ，你插5至7 - S = [5 6 7 8] 。以此方式，替换步骤选自S的元素仅仅复制到A [K + 1：N-1]

This is not needed. You can try to keep the S array sorted all the time. Every time you want to insert a new number in the array, you insert it in such, so the array could stay sorted. For example, if you have S = [5 7 8] so far and you want to insert 6, you inserted between 5 and 7 - S = [5 6 7 8]. This way the replacement step is just copying the elements from S to A[k+1:N-1].