# 172 - 151 翻转字符串里面的单词 - 044

## 题目

给定一个字符串，逐个翻转字符串中的每个单词。

示例 1：

> 输入: "the sky is blue" 输出: "blue is sky the"

示例 2：

> 输入: " hello world! " 输出: "world! hello" 解释: 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。

示例 3：

> 输入: "a good example" 输出: "example good a" 解释: 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。

说明：

* 无空格字符构成一个单词。
* 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。
* 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。

## 解答

1. 字符串进来后前后去掉空格
2. 空格拆开
3. 翻转字符串，用空格连起来

   如果一个字符是空，那么就跳过

```python
class Solution:
    def reverseWords(self, s: str) -> str:
        s = s.strip()
        words = s.split(" ")
        return " ".join([word for word in reversed(words) if word != ""])
```

> Runtime: 32 ms, faster than 75.81% of Python3 online submissions for Reverse Words in a String.
>
> Memory Usage: 13.3 MB, less than 100.00% of Python3 online submissions for Reverse Words in a String.

```javascript
var reverseWords = function(s) {
  s = s.trim().split(" ").reverse()
  let ans = ""
  for (const item of s) {
    if (item) {
      ans += item + " "
    }
  }
  return ans.slice(0, ans.length - 1)
};
```

> Runtime: 72 ms, faster than 13.52% of JavaScript online submissions for Reverse Words in a String.
>
> Memory Usage: 34.4 MB, less than 90.91% of JavaScript online submissions for Reverse Words in a String.

```go
func reverseWords(s string) string {
    s = strings.Trim(s, " ")
    if s == "" {
        return s
    }
    words := strings.Split(s, " ")
    var ans string
    for i := len(words) - 1; i >= 0; i-- {
        if words[i] != "" {
            ans += words[i] + " "
        }
    }
    return ans[0 : len(ans)-1]
}
```

> Runtime: 4 ms, faster than 49.48% of Go online submissions for Reverse Words in a String.
>
> Memory Usage: 8.4 MB, less than 50.00% of Go online submissions for Reverse Words in a String.


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