1 # manner a cycle
2 temStr = ' adbeorj1305 '
. 3 newStr = ''
. 4 I = 0
. 5 the while I < len (temStr):
. 6 newStr + = newStr temStr [len (temStr)-l- I]
. 7 I = I + . 1
. 8 Print (newStr, ' string inversion ' )
. 9
10 # way two sections
. 11 temStr = ' adbeorj1305 '
12 is Print (temStr [:: -. 1], ' slice ' )
13 is
14 #Three ways to use the reverse method list
15 temStr = ' adbeorj1305 '
16 Ll = List (temStr)
. 17 L1.reverse ()
18 is S2 = '' .join (Ll)
. 19 Print (S2, " borrowing inversion method list ' )
1 # Four ways the lambda using the reduce
2 #lambda used to write a simple function, but def more powerful to handle tasks. Which way it clearer which way use, do not blindly use lambda expressions.
. 3
. 4 # manner using five stacks
. 5 temStr = ' adbeorj1305 '
. 6 Ll = List (temStr)
. 7 Sl = ''
. 8 the while len (Ll)> 0:
. 9 Sl = Sl + L1.pop ()
10 Print (Sl, ' using stack mode using pop ' )
11 # ###################### following is an example of failure
12 is temStr = ' adbeorj1305 '
13 is Ll = List (temStr)
14 = Sl ''
15 for I in Ll:
16 Sl = Sl + L1.pop () # also after sequentially pop, stitching
. 17 Print (Sl, ' Ll ' , Ll)
18 is # Print (Sl, 'the use of the stack, using pop mode', Ll)
. 19 # # output They are: 5031jr why the latter is not performed, (⊙ o ⊙) ...,
20 is ' ''
21 is Causes: pop after execution, the variable length L1, and the index value i in L1 accessed in increasing i,
22 when i = r, index = 6 enters the next cycle,
23 but this time L1 = [ 'a', ' d', 'b', 'e', 'o'] =. 5 the maximum index,
24 thus cycle end, subsequent elements can not be added.
25 '' '