url: https://leetcode.com/problems/rotate-array/
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?
解决方案
/** * 一个数组,比如nums=[1,2,3]: * 如果rotate的次数k=其长度自身,那么数组不变 ([3,1,2], [2,3,1], [1,2,3]); * 如果rotate的次数k>其长度自身,那么数组的“有效”rotate次数是k%nums.length; * 如果rotate的次数k<其长度自身,先把要移到前面的放到另外的空间,剩余元素往后移动,再把之前移到另外空间的元素转移回来。 */ class Solution { public void rotate(int[] nums, int k) { if(nums == null || nums.length < 1){ return; } //计算有效rotate次数 if(k > nums.length){ k = k % nums.length; } //直接返回 if(k == nums.length){ return; } int[] remain = new int[k]; int j = 0; //先把要移到前面的放到另外的空间 for(int i = nums.length - k; i < nums.length; ++i){ remain[j] = nums[i]; ++j; } //剩余元素往后移动 for(int q = nums.length - k - 1; q >= 0; --q){ nums[q+k] = nums[q]; } //再把之前移到另外空间的元素转移回来。 for(int p = 0; p < remain.length; ++p){ nums[p] = remain[p]; } } }