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