[Leetcode]344. Reverse String

📄 Description

Write a function that reverses a string. The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

🔨 My Solution

• use two pointers for exchanging elements

💻 My Submission

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        left=0
        right=len(s)-1
        while left<=right:
            s[left],s[right]=s[right],s[left]
            left+=1
            right-=1

🔎 Complexity

Time Complexity: $O(n)$
Space Complexity: $O(1)$

💡 Other Approches

📌 Pythonic Way(파이썬 다운 방식)

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s.reverse()

📌 Using Slicing(슬라이싱)

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:]=s[::-1]

🔴 공간 복잡도가 $O(1)$로 제한되어 있으므로 s=s[::-1]은 통과되지 않는다.

💭 s[:]=s[::-1] 랑 s=s[::-1] 가 어떻게 다른데?

s = s[::-1] creates a new local variable s and assigns the reversed list to it, it doesn't modify the list s, whereas s[:] = s[::-1] creates an intermediate reversed list, and assigns it back to s , since we are using slice notation s[:].

References

• https://leetcode.com/problems/reverse-string/
• https://leetcode.com/problems/reverse-string/discuss/80946/Python2.7-(3-solutions%3A-Recursive-Classic-Pythonic)
• 파이썬 알고리즘 인터뷰(95가지 알고리즘 문제 풀이로 완성하는 코딩 테스트), 박상길 지음