C++实现LeetCode(557.翻转字符串中的单词之三)

• [LeetCode] 557.Reverse Words in a String III 翻转字符串中的单词之三

[LeetCode] 557.Reverse Words in a String III 翻转字符串中的单词之三

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：

 class Solution { 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：

 class Solution { 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 j=end - 1; j; ++i, --j) swap(s[i], s[j]); } start=++end; return s; };

类似题目：

Reverse Words in a String II

Reverse Words in a String

参考资料：

https:来源gao@!dai!ma.com搞$$代^@码!网//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