题目
-
编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。
不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
示例 1：
输入：s = ["h","e","l","l","o"]
输出：["o","l","l","e","h"]
示例 2：
输入：s = ["H","a","n","n","a","h"]
输出：["h","a","n","n","a","H"]
法一、库函数
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
return s.reverse()
- O(1) 的额外空间
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:]=s[::-1]
- O(N) 的额外空间。切片操作会创建一个新的列表来存储反转后的字符串。即使最终将反转后的字符串赋值回原始列表s
法二、双指针
- 思想：i指针指向第一个元素，j指针指向最后一个元素，只要i,j指针没有相遇，不断交换i,j指针所指两个元素
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
n = len(s)
i,j = 0,n-1
while(i<j):
tmp = s[i]
s[i] = s[j]
s[j] = tmp
i+=1
j-=1
return s