[leetcode][array] Rotate Array

题目;

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.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?

Related problem: Reverse Words in a String II

方法一：

class Solution {public://最直观的解法：循环移动k次，每次移动n个元素 O(n*k) 超时void rotate(vector<int>& nums, int k) {if (nums.size() <= 1) return;int n = nums.size();k = k%n;for (int i = 0; i < k; ++i){int cur = nums[n - 1];//暂存最后一个元素for (int j = n - 1; j > 0; --j){nums[j] = nums[j - 1];}nums[0] = cur;}}};

方法二：

class Solution {public://时间复杂度O(n),空间复杂度O(k)void rotate(vector<int>& nums, int k) {if (nums.size() <= 1) return;int n = nums.size();k = k%n;//暂存后半部分vector<int> tmp(nums.end() - k, nums.end());//前半部分整体后移k步for (int i = n - 1; i >= k; --i){nums[i] = nums[i - k];}//将暂存的后半部分放入前半部分for (int i = 0; i < k; ++i){nums[i] = tmp[i];}}};

注：本方法还可以做一下空间的优化：不一定暂存后半部分，可以暂存元素较少的那一部分。

方法三：

class Solution {public://先分别反转前后两部分，然后反转整个数组 时间复杂度O(n),空间复杂度O(1)void rotate(vector<int>& nums, int k) {if (nums.size() <= 1) return;int n = nums.size();k = k%n;reverse(nums.begin(), nums.end() - k);reverse(nums.end() - k, nums.end());reverse(nums.begin(), nums.end());}};