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

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 = start, j = end - 1; i < j; ++i, --j) {

									                swap(s[i], s[j]);

									            }

									            start = ++end;

									        }

									        return s;

									    }

									};