题目

https://leetcode-cn.com/problems/reverse-words-in-a-string-iii/

给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。

示例 1:

输入: "Let's take LeetCode contest"
输出: "s'teL ekat edoCteeL tsetnoc" 

注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。

我的AC

class Solution(object):
    def reverseWords(self, s):
        """
        :type s: str
        :rtype: str
        """
        length = len(s)
        word_list = s.split()
        rev_list = [word[::-1] for word in word_list]
        rev_str = " ".join(rev_list)
        return rev_str

小结

s.split() # 字符串分割
rev_str = " ".join(rev_list) # 列表元素拼接为字符串

完整的一个 code

class Solution:
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        result = []
        for i in nums1:
            for j in nums2:
                if i == j:
                    result.append(i) # 插入交集
                    nums2.remove(j)  # 插入过的值不再出现
                    break
        return result
    
if __name__ == "__main__":
    nums1 = [1,2,2,1]
    nums2 = [1,2]
    s = Solution()
    print(s.intersect(nums1, nums2))

原文：https://blog.csdn.net/qq_36190978/article/details/87286512