189. Rotate Array

最新推荐文章于 2024-01-03 12:46:00 发布

转载最新推荐文章于 2024-01-03 12:46:00 发布 · 45 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/anghostcici/p/6917120.html

本文介绍了一种将数组元素向右旋转k步的操作，并提供了两种解决方案。第一种方法通过不断将数组的第一个元素移到末尾来实现旋转，但时间复杂度较高。第二种方法采用三次反转的方式，先反转前n-k个元素，再反转后k个元素，最后反转整个数组，实现了O(n)的时间复杂度。

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.

Solution 1: note that line 6 uses k%size. the time complexity of erase is O(n), so the worst total time could be O(n^2)

 1 class Solution {
 2 public:
 3     void rotate(vector<int>& nums, int k) {
 4         int size=nums.size();
 5         int i=0;
 6         while (i<size-k%size){
 7             nums.push_back(nums[0]);
 8             nums.erase(nums.begin());
 9             i++;
10         }
11     }
12 };

Solution 2:reverse the previous n-k nums and then reverse the later k nums. Finally reverse all. O(n)

1 2 3 4 5 6 7

4321567

4 3 2 1 7 6 5

5 6 7 1 2 3 4

 1 class Solution {
 2 public:
 3     void rotate(vector<int>& nums, int k) {
 4         if (nums.empty() || (k %= nums.size()) == 0) return;
 5         int n = nums.size();
 6         reverse(nums.begin(), nums.begin() + n - k);
 7         reverse(nums.begin() + n - k, nums.end());
 8         reverse(nums.begin(), nums.end());
 9     }
10 };

转载于:https://www.cnblogs.com/anghostcici/p/6917120.html