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"]

 

Constraints:

• 1 <= s.length <= 105
• s[i] is a printable ascii character.

Solutions

Solution 1: Two Pointers

We use two pointers $i$ and $j$, initially pointing to the start and end of the array respectively. Each time, we swap the elements at $i$ and $j$, then move $i$ forward and $j$ backward, until $i$ and $j$ meet.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def reverseString(self, s: List[str]) -> None:
        i, j = 0, len(s) - 1
        while i < j:
            s[i], s[j] = s[j], s[i]
            i, j = i + 1, j - 1

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

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

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

/**
 Do not return anything, modify s in-place instead.
 */
function reverseString(s: string[]): void {
    for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
        [s[i], s[j]] = [s[j], s[i]];
    }
}

impl Solution {
    pub fn reverse_string(s: &mut Vec<char>) {
        let mut i = 0;
        let mut j = s.len() - 1;
        while i < j {
            s.swap(i, j);
            i += 1;
            j -= 1;
        }
    }
}

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

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