巧妙变换

题目分析来自“待字闺中”微博公众平台。

一个数组，arr = ['a1', 'a2', 'a3', 'a4', 'b1', 'b2', 'b3', 'b4']，变换为：arr = ['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']

第一次使用Python编写程序，肯定有许多不妥的地方。

第一种方法：

__author__ = 'ledao'

arr = ['a1', 'a2', 'a3', 'a4', 'b1', 'b2', 'b3', 'b4']

length = len(arr)
halfLen = length/2
counter = halfLen-1

for i in xrange(halfLen, length-1):

    local_counter = counter
    counter -= 1
    if -1 == counter:
        break
    local_index = i
    while local_counter > 0:
        val = arr[local_index]
        arr[local_index] = arr[local_index-1]
        arr[local_index-1] = val
        local_index -= 1
        local_counter -= 1

print arr

 第二种方法：

__author__ = 'ledao'

arr = ['a1', 'a2', 'a3', 'a4', 'a5', 'b1', 'b2', 'b3', 'b4', 'b5']

length = len(arr)
halflen = length/2

for i in xrange(1, halflen):
    for j in xrange(0, i):
        local_val = arr[halflen+i-1-2*j]
        arr[halflen+i-1-2*j] = arr[halflen+i-1-2*j-1]
        arr[halflen+i-1-2*j-1] = local_val

print arr

 第三种方法：

__author__ = 'ledao'

arr = ['a1', 'a2', 'a3', 'a4', 'a5', 'b1', 'b2', 'b3', 'b4', 'b5']

def finalIndex(old_index, halflen):
    return (2*old_index)%(2*halflen-1)

length = len(arr)
modified = ['False']*length

for i in xrange(0,length-1):
    if modified[i] == 'False':
        firstIndex = i
        stored_val = arr[firstIndex]
        while True:
            local_value = stored_val
            if modified[firstIndex] == 'False':
                modified[firstIndex] = 'True'
            else:
                break;
            local_finalIndex = finalIndex(firstIndex, length/2)
            stored_val = arr[local_finalIndex]
            arr[local_finalIndex] = local_value
            firstIndex = local_finalIndex

print arr

 第四种方法由于水平有限，没看懂……T-T，讲解连接如下，请教高手！

http://www.newsmth.net/bbscon.php?bid=1032&id=47005