Rotate Array(leetCode189)
阿新 • • 發佈:2018-11-26
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?
//使用雙端佇列 想怎麼幹怎麼幹 和利用的空間比較多 public static int[] rotate(int[] nums, int k) { // ArrayDeque, ConcurrentLinkedDeque, LinkedBlockingDeque, LinkedList Deque deque = new ArrayDeque(); for (int num:nums){ deque.add(num); } for (int j = 0;j<k;j++){ int last = (int) deque.getLast(); deque.removeLast(); deque.addFirst(last); } int m = 0; while (deque.peek()!= null){ nums[m] = (int) deque.pop(); m++; } System.out.println(Arrays.toString(nums)); return nums; } public static int[] rotate2(int[] nums, int k) { k = k % nums.length; int[] kHolder = new int[k]; System.arraycopy(nums, nums.length - k, kHolder, 0, k); System.arraycopy(nums, 0, nums, k, nums.length - k); System.arraycopy(kHolder, 0, nums, 0, k); return kHolder; } public static int[] rotate3(int[] nums, int k) { if(k==0 || nums.length <2) { return nums; } int len = nums.length; int r = k % len; int[] temp = new int[len]; for(int j=0; j<len; j++){ temp[(j+r)%len] = nums[j]; } System.arraycopy(temp, 0, nums, 0, temp.length); return temp; }
題目地址:https://leetcode.com/problems/rotate-array/