[Leetcode]344. Reverse String

limelimejiwon·2022년 3월 11일
0

📄 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)O(n)
Space Complexity: O(1)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)O(1)로 제한되어 있으므로 s=s[::-1]은 통과되지 않는다.

💭 s[:]=s[::-1]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

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