Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
Example 2:
Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Could you do it in-place with O(1) extra space?
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Approach 1: Using Extra Array
classSolution {publicvoidrotate(int[] nums,int k) {int[] a =newint[nums.length];for (int i =0; i <nums.length; i++) { a[(i + k) %nums.length] = nums[i]; }for (int i =0; i <nums.length; i++) { nums[i] = a[i]; } }}
Approach #2 Using Cyclic Replacements Confusion
classSolution {publicvoidrotate(int[] nums,int k) { k = k %nums.length;int count =0;for (int start =0; count <nums.length; start++) {int current = start;int prev = nums[start];do {int next = (current + k) %nums.length;int temp = nums[next]; nums[next] = prev; prev = temp; current = next; count++; } while(start != current); } }}