【剑指offer】面试题42：单词翻转顺序&左右旋转字符串

这里尽可能的不去用语言本身提供的函数。

将string逆置

def reverse(string):
#return string[::-1]
reversedStr = ''
for i in xrange(len(string) - 1, -1, -1):
reversedStr += string[i]
return reversedStr

单词的翻转

def reverseWords(string):
if len( string ) <= 1: return
split = ' '
reversedStr = ''
word = ''
# reverse every word in string
for c in string:
if c == split:
word = reverse(word) + split
reversedStr += word
word = ''
else:
word += c
# the last word
reversedStr += reverse(word)
# reverse the whole string
reversedStr = reverse(reversedStr)
return reversedStr

这里没有考虑前导空格之类的复杂情况，默认string是非常规整的，单词间由一个空格隔开。

字符串的左右旋转，右旋转我们将字符串后面的n个字符移到最前面。通过参数direction来区别，实参为‘l'，’L‘， ’r‘， ’R'.

'''
@ if direction = 'l' or 'L', move n items from front to end
@ if direction = 'r' or 'R', move n items from back to front
'''
def RotateString(string, n, direction):
if n >= len(string) or n <= 0:
return string
reversedStr = ''
leftpart = ''
rightpart = ''

# move last n chiracter to front
if direction == 'r' or direction == 'R':
n = len(string) - n

for i in range(n):
leftpart += string[i]

for i in range(n, len(string)):
rightpart += string[i]

reversedStr += reverse(leftpart)
reversedStr += reverse(rightpart)
reversedStr = reverse(reversedStr)
return reversedStr

若觉得单词翻转的实现繁琐，也可以直接用语言本身提供的函数，但是不推荐。

def reverseWords(string):
return ' '.join(string.split()[::-1])