[Swift]LeetCode151. 翻转字符串里的单词 | Reverse Words in a String

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

Example:  

Input: "",
Output: "".
the sky is blueblue is sky the

Note:

• A word is defined as a sequence of non-space characters.
• Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
• You need to reduce multiple spaces between two words to a single space in the reversed string.

Follow up: For C programmers, try to solve it in-place in O(1) space.

---

给定一个字符串，逐个翻转字符串中的每一个单词。

示例:  

输入: "",
输出: "".
the sky is blueblue is sky the

说明:

• 无空格字符构成一个单词。
• 输入字符串能够在前面或者后面包含多余的空格，可是反转后的字符不能包括。
• 若是两个单词间有多余的空格，将反转后单词间的空格减小到只含一个。

进阶: 请选用C语言的用户尝试使用 O(1) 空间复杂度的原地解法。

---

1 class class Solution {
2     func reverseWords(_ s: String) -> String {
3           return String(s.split(separator: " ").reversed().reduce("") { total, word in total + word + " "}.dropLast())
4     }
5 }