My accepted Java solution O(n) time and O(1)space

  • 8
    public class Solution {
        public void rotate(int[] nums, int k) {
           int end = nums.length;
            k=k%end;// to check for outofbounds when k >= nums.length
            rotate(nums,0,end-k-1);//reverse one half of the array
            rotate(nums,end-k,end-1);//reverse other half of the array
            rotate(nums,0,end-1);//reverse whole array  
        public void swap(int[] nums,int a,int b){
            int temp = nums[a];
            nums[a] = nums[b];
            nums[b] = temp;
        public void rotate(int[] nums,int start,int end){
            for(int i = start;i<=(start+end)/2;i++){

  • 0

    Super clear O(1) solution!

    Reverse 0...l-k part and reverse l-k...l part.
    After that , the letters are reversed in each part but the two part is in original order.
    Then reverse all.
    Both reversed parts restored but the order of two part is reversed.

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