题解 | #字符串变形#
字符串变形
https://www.nowcoder.com/practice/c3120c1c1bc44ad986259c0cf0f0b80e
class Solution: def transform(self, s: str): res = "" for i in s: if 'A' <= i <= 'Z': res += chr(ord(i) + 32) elif 'a' <= i <= 'z': res += chr(ord(i) - 32) return res def trans(self , s: str, n: int) -> str: # write code here tmp = s.split(" ") print(tmp) i = 0 j = len(tmp)-1 while i <= j: tmp[i], tmp[j] = self.transform(tmp[j]), self.transform(tmp[i]) i += 1 j -= 1 res = " ".join(tmp) return res
- 先将字符串以空格为分隔符进行切分;
- 然后设置双指针,一个指向首端一个指向尾端,向中间靠拢遍历字符串数组;
- 将双指针指向的元素值先做大小写处理,再进行交换;
- 最后将字符串数组以空格的形式进行连接。