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"]

Constraints:

1 <= s.length <= 10^5
s[i] is a printable ascii character.

Approach 1: In-place modification

We iterate the array to process two elements at the same time and swap them. First we swap s[0] and s[n - 1]. Then swap s[1] and s[n - 2] and so on. Therefore, we only swap n / 2 times.

class Solution {
public:
    void reverseString(vector<char>& s) {
        int n = (int) s.size();
        for (int i = 0; i < n / 2; i++) {
            swap(s[i], s[n - 1 - i]);
        }
    }
};

func reverseString(s []byte)  {
    i, j := 0, len(s) - 1
    for i < j {
        s[i], s[j] = s[j], s[i]
        i += 1
        j -= 1
    }
}

class Solution {
    public void reverseString(char[] s) {
        int i = 0, j = s.length - 1;
        while (i < j) {
            char c = s[i];
            s[i++] = s[j];
            s[j--] = c;
        }
    }
}

/**
 * @param {character[]} s
 * @return {void} Do not return anything, modify s in-place instead.
 */
var reverseString = function (s) {
  let left = 0;
  let right = s.length - 1;
  while (left < right) {
    char = s[left];
    s[left] = s[right];
    s[right] = char;
    left++;
    right--;
  }
};

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

        left, right = 0, len(s) - 1
        while left < right:
            s[left], s[right] = s[right], s[left]
            left += 1
            right -= 1

Approach 2: STL

class Solution {
public:
    void reverseString(vector<char>& s) {
        reverse(s.begin(), s.end());
    }
};

class Solution:
    def reverseString(self, s: List[str]) -> None:
        s.reverse()

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
       s.reverse()
    }
}

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
       s.reverse()
    }
}

0344 - Reverse String (Easy)

Problem Link

Problem Statement

Approach 1: In-place modification

Approach 2: STL

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