剑指 Offer 58 - II. 左旋转字符串

class Solution:
    # 切片法
    # def reverseLeftWords(self, s: str, n: int) -> str:
    #     # s:
    #     temp = s[:n]
    #     s = s[n:]
    #     s = s + temp
    #     return s
    # def reverseLeftWords(self, s: str, n: int) -> str:
    #     return s[n:]+s[:n]
    # 列表遍历拼接
    # def reverseLeftWords(self, s: str, n: int) -> str:
    #     temp = []
    #     for i in range(n, len(s)):
    #         temp.append(s[i])
    #     for i in range(0, n):
    #         temp.append(s[i])
    # # string.join(seq)：以string作为分隔符，将seq中所有的字符以string分隔开
    # # 换一种思路，如果string为空，那么就是将seq中所有的字符连接起来。
    # # seq可以是列表
    #     return ''.join(temp)
    # 字符串遍历拼接
     def reverseLeftWords(self, s: str, n: int) -> str:
        temp = ''
        for i in range(n, len(s)):
            temp += s[i]
        for i in range(0, n):
            temp += s[i]
        return temp
    # 效率比较 切片>列表遍历拼接>字符串遍历拼接
    # os：虽然字符串遍历拼接是最熟悉的但是效率却是最低下的。。