




Sponsored
Sponsored
This approach involves splitting the given string into words, reversing each word individually, and then joining them back together.
Time Complexity: O(n), where n is the length of the string.
Space Complexity: O(1), as only a constant amount of extra space is used.
1public class Solution {
2    public static void reverseWords(String s) {
3        char[] arr = s.toCharArray();
4        int n = arr.length;
5        int start = 0;
6
7        for (int end = 0; end < n; end++) {
8            if (arr[end] == ' ' || end == n - 1) {
9                reverse(arr, start, (end == n - 1) ? end : end - 1);
10                start = end + 1;
11            }
12        }
13        System.out.println(new String(arr));
14    }
15
16    private static void reverse(char[] arr, int start, int end) {
17        while (start < end) {
18            char temp = arr[start];
19            arr[start] = arr[end];
20            arr[end] = temp;
21            start++;
22            end--;
23        }
24    }
25
26    public static void main(String[] args) {
27        String s = "Let's take LeetCode contest";
28        reverseWords(s);
29    }
30}This Java solution uses a similar approach to the C solution. It reverses each word in place using a helper function.
In this C solution, a two-pointer technique is applied to reverse individual words within the string in place, using a helper function.