为有牺牲多壮志,敢教日月换新天。

[Swift]LeetCode434. 字符串中的单词数 | Number of Segments in a String

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/9783619.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters.

Please note that the string does not contain any non-printable characters.

Example:

Input: "Hello, my name is John"
Output: 5

统计字符串中的单词个数,这里的单词指的是连续的不是空格的字符。

请注意,你可以假定字符串里不包括任何不可打印的字符。

示例:

输入: "Hello, my name is John"
输出: 5

8ms
 1 class Solution {
 2     func countSegments(_ s: String) -> Int {
 3         if s == nil || s.count == 0 {return 0}
 4         var word:Int = 0
 5         var count:Int = 0
 6         //按索引遍历
 7         for i in s.indices
 8         {
 9             if s[i] == " "
10             {
11                 if word > 0
12                 {
13                     count += 1
14                     word = 0
15                 }
16             }
17             else
18             {
19                 word += 1
20             }            
21         }
22         if word > 0
23         {
24             count += 1
25         }
26         return count
27     }
28 }

8ms

 1 class Solution {
 2     func countSegments(_ s: String) -> Int {
 3         var count = 0
 4         var segmentInProgress = false
 5         for char in s {
 6             if char != " " && !segmentInProgress {
 7                 count += 1
 8                 segmentInProgress = true
 9             } else if char == " " {
10                 segmentInProgress = false
11             }
12         }
13         
14         return count
15     }
16 }

12ms

1 class Solution {
2     func countSegments(_ s: String) -> Int {
3         return s.split(separator: " ").count
4     }
5 }

16ms

1 class Solution {
2     func countSegments(_ s: String) -> Int {
3         return s.characters.split(separator: " ").count
4     }
5 }

 

posted @ 2018-10-13 17:50  为敢技术  阅读(356)  评论(0编辑  收藏  举报