使用reverse來讓列表反轉(zhuǎn)特別方便, 沒事自己寫了幾種方式

In [59]: def reverse(nums):

length = len(nums)

for i in range(length-1):

nums.insert(length-i-1, nums.pop(0))

print nums

....:

In [60]:

In [60]:

In [60]: a = range(5)

In [61]: reverse(a)

[1, 2, 3, 4, 0]

[2, 3, 4, 1, 0]

[3, 4, 2, 1, 0]

[4, 3, 2, 1, 0]

In [62]: a

Out[62]: [4, 3, 2, 1, 0]

In [63]:

根據(jù)結(jié)果能看出, insert操作的執(zhí)行順序是先刪除, 后插入

In [67]: def reverse(nums):

i, j = 0, len(nums)-1

while i < j:

nums[i], nums[j] = nums[j], nums[i]

i, j = i+1, j-1

print nums

In [69]: a = range(5)

In [70]: a

Out[70]: [0, 1, 2, 3, 4]

In [71]:

In [71]: reverse(a)

[4, 1, 2, 3, 0]

[4, 3, 2, 1, 0]

In [72]: a

Out[72]: [4, 3, 2, 1, 0]

In [73]:

以上兩種都是原地操作

In [78]: a

Out[78]: [0, 1, 2, 3, 4]

In [79]:

In [79]: b = []

In [80]: def reverse(nums):

....: for i in nums:

....: b.insert(0, i)

....:

In [81]:

In [81]: reverse(a)

In [82]:

In [82]: b

Out[82]: [4, 3, 2, 1, 0]

In [83]: id(a)

Out[83]: 140434157093160

In [84]: id(b)

Out[84]: 140434157062984

In [85]:

總結(jié)

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