[Leetcode]Reverse Words in a String

Given an input string, reverse the string word by word.

For example,

Given s = "

the sky is blue

",

return "

blue is sky the

".

Clarification:

What constitutes a word?

A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?

Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?

Reduce them to a single space in the reversed string.

把字符串里的单词反转过来，注意要去掉多余的空格~最直接的方法是把字符串按空格分开，把得到的单词存到数组里，然后把单词反转过来得到结果~如果直接调用python的split()函数，代码就很简洁，如下~

class Solution:
# @param s, a string
# @return a string
def reverseWords(self, s):
if s is None or len(s) == 0: return ""
return ' '.join(s.split()[::-1])

也可以不调用split()等这些内置的函数~从后往前扫描字符串，记录每个单词的起点和终点位置，加到数组words中，然后再转换成要求的字符串~

class Solution:
# @param s, a string
# @return a string
def reverseWords(self, s):
if s is None or len(s) == 0: return ""
words = []; res = ''
i = len(s) - 1
while i >= 0:
if s[i] == ' ':
i -= 1
else:
end = i; start = end - 1
while start >= 0 and s[start] != ' ':
start -= 1
words.append(s[start + 1: end + 1])
i = start
for i in xrange(len(words)):
if i != len(words) - 1:
res += words[i] + ' '
else:
res += words[i]
return res这题还有另外一种思路，先把整个字符串反转并去掉多余的空格，然后再对里面的每一个单词进行反转~