Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note: Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
旋转数组
解法一:
使用交换数据的方式,对于上面n=7,k=3的情况,可以分成三步完成:
将操作分成三步即可: 时间复杂度是O(N),,空间复杂度是O(1)
class Solution {public://解法1:使用交换void rotate(vector<int>& nums, int k) {int length=nums.size();k=k%length;swapVector(nums,0,length-k-1);swapVector(nums,length-k,length-1);swapVector(nums,0,length-1);}void swapVector(vector<int> &nums,int first,int last){for(int i=first,j=last;i<j;i++,j–){swap(nums[i],nums[j]);}}};解法二:
直接使用一个新的数组来保存数据,再将新的数组复制给给定数组。 时间复杂度是O(n),空间复杂度是O(N)。
class Solution {public://直接使用一块新的内存void rotate(vector<int>& nums, int k) {vector<int> result;int length=nums.size();k%=length;result.insert(result.end(),nums.begin()+length-k,nums.end());result.insert(result.end(),nums.begin(),nums.begin()+length-k);nums=result;}};
我们可以沿途用镜头记录彼此的笑脸,和属于我们的风景。