Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
Example 1:
Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"
Note: In the string, each word is separated by single space and there will not be any extra space in the string.
这道题让我们翻转字符串中的每个单词，感觉整体难度要比之前两道Reverse Words in a String II和Reverse Words in a String要小一些，由于题目中说明了没有多余空格，使得难度进一步的降低了。首先我们来看使用字符流处理类stringstream来做的方法，相当简单，就是按顺序读入每个单词进行翻转即可，参见代码如下：
解法一：
public: string reverseWords(string s) { string res = "", t = ""; istringstream is(s); while (is >> t) { reverse(t.begin(), t.end()); res += t + " "; } res.pop_back(); return res; } };
下面我们来看不使用字符流处理类，也不使用STL内置的reverse函数的方法，那么就是用两个指针，分别指向每个单词的开头和结尾位置，确定了单词的首尾位置后，再用两个指针对单词进行首尾交换即可，有点像验证回文字符串的方法，参见代码如下：
解法二：
public: string reverseWords(string s) { int start = 0, end = 0, n = s.size(); while (start < n && end < n) { while (end < n && s[end] != ' ') ++end; for (int i = start, j = end - 1; i < j; ++i, --j) { swap(s[i], s[j]); } start = ++end; } return s; } };
参考资料：
https://discuss.leetcode.com/topic/85773/nothing-fancy-straight-java-stringbuilder
https://discuss.leetcode.com/topic/85797/java-two-methods-3-line-using-built-in-and-char-array
本文转自博客园Grandyang的博客，原文链接：[LeetCode] Reverse Words in a String III 翻转字符串中的单词之三
，如需转载请自行联系原博主。
