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];
        }
    }
}