旋转使用LINQ语法定义的数组

问题描述

我解决旋转阵列的这个问题，并得到了算法和code工作

I am solving this problem of rotating an array and got algorithm and code working

 int[] Rotate(int[] ar,int k)
        {
            if (k <= 0 || k > ar.Length - 1)
                return ar;
            Reverse(ar, 0, k - 1);
            Reverse(ar, k, ar.Length - 1);
            Reverse(ar, 0, ar.Length - 1);
            return ar;
        }

 void Reverse(int[] ar,int start, int end)
        {
            while (start < end)
            {
                int temp = ar[start];
                ar[start] = ar[end];
                ar[end] = temp;
                start++;
                end--;
            }
        }

现在我想这样做LINQ中，我得到了下面的code，我觉得这是可以做到更好。

Now I want to do this in LINQ and I got the below code, I think this can be done much better.

 int[] Rotate(int[] ar,int k)
    {
        if (k <= 0 || k > ar.Length - 1)
            return ar;
        int[] ar1=ar.Take(k-1).Reverse().ToArray();
        int[] ar2=ar.Skip(k - 1).Take(ar.Length - k+1).Reverse().ToArray();
        int[] ar3 = ar1.Concat(ar2).Reverse().ToArray();
        return ar3;
    }

这是从编程珍珠一个众所周知的算法 - <一个href=\"http://books.google.com/books?id=kse_7qbWbjsC&lpg=PA14&ots=DfzTzQCSar&dq=rotate%20an%20array%20programming%20pearls&pg=PA14#v=onepage&q&f=false\" rel=\"nofollow\">http://books.google.com/books?id=kse_7qbWbjsC&lpg=PA14&ots=DfzTzQCSar&dq=rotate%20an%20array%20programming%20pearls&pg=PA14#v=onepage&q&f=false

This is a well known algorithm from Programming pearls - http://books.google.com/books?id=kse_7qbWbjsC&lpg=PA14&ots=DfzTzQCSar&dq=rotate%20an%20array%20programming%20pearls&pg=PA14#v=onepage&q&f=false

而在一般情况如何发展我的LINQ的技能，如果我给出规划问题，现在我只想在循环或foreach循环，怎么想的LINQ运营商而言。我读C＃4.0概括地说，比任何练习其他的建议？

And in general how to develop my LINQ skills, if I am given a programming problem, right now I am only thinking in for loops or foreach loops, how to think in terms of linq operators. I am reading C# 4.0 nutshell, other than practicing any advice?

推荐答案

与code开始：

int[] ar1=ar.Take(k-1).Reverse().ToArray();
int[] ar2=ar.Skip(k - 1).Take(ar.Length - k+1).Reverse().ToArray();
int[] ar3 = ar1.Concat(ar2).Reverse().ToArray();

既然你只是想获得的所有剩余的元素，是不是需要在第二行取。

Since you just want to get all of the remaining elements, the Take in the second line isn't needed.

Ar1和Ar2只是列举，所以它们不需要是数组。不需要的ToArray的呼叫。有了一点创造性重命名的抛出，我们有：

ar1 and ar2 are just enumerated, so they don't need to be arrays. The ToArray calls aren't needed. With a bit of creative renaming thrown in, we have:

IEnumerable<int> revFirst = ar.Take(k-1).Reverse();
IEnumerable<int> revLast = ar.Skip(k-1).Reverse();
int[] ar3 = revFirst.Concat(revLast).Reverse().ToArray();

现在我们

转（REV（第一）+ REV（最后））

rev ( rev(first) + rev(last) )

分发外转给

转（REV（最后））+转（REV（第一））

rev ( rev(last) ) + rev ( rev(first) )

这是相同的

last + first

施加相同的操作到code给出

applying the same operations to the code gives

IEnumerable<int> first = ar.Take(k-1);
IEnumerable<int> last = ar.Skip(k-1);
int[] ar3 = last.Concat(first).ToArray();

进一步简化，以

int[] ar3 = ar.Skip(k-1).Concat(ar.Take(k-1)).ToArray();

现在我们已经乔恩斯基特的答案，所以我们必须做到的。

and now we have Jon Skeet's answer so we must be done.

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