Write a function that reverses a string. The input string is given as an array of characters s
.
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.
Follow up: Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1)
extra memory.
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-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; ++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]
}
}
/**
* @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]];
}
};