189. Rotate Arrayeasy

Posted 2020-10-07 abc_begin

189. Rotate Array【easy】

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

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

 

 解法一：

 1 class Solution {
 2 public:
 3     void rotate(vector<int>& nums, int k) {
 4         int len = nums.size();
 5         
 6         if (len == 0 || k % len == 0)
 7         {
 8             return;
 9         }
10         
11         int mid = len - k % len;
12         
13         reverseArray(nums, 0, mid - 1);
14         reverseArray(nums, mid, len - 1);
15         reverseArray(nums, 0, len - 1);
16     }
17     
18     void reverseArray(vector<int>& nums, int start, int end)
19     {
20         int s = start;
21         int e = end;
22         
23         while (s <= e)
24         {
25             int temp = nums[s];
26             nums[s] = nums[e];
27             nums[e] = temp;
28             s++;
29             e--;
30         }
31     }
32 };

注意：

1、边界值检查，避免对0取余和长度不合法

2、先分别翻转，再总体翻转，注意下标

 

解法二：

 1 class Solution 
 2     {
 3     public:
 4         void rotate(int nums[], int n, int k) 
 5         {
 6             k = k%n;
 7     
 8             // Reverse the first n - k numbers.
 9             // Index i (0 <= i < n - k) becomes n - k - i.
10             reverse(nums, nums + n - k);
11             
12             // Reverse tha last k numbers.
13             // Index n - k + i (0 <= i < k) becomes n - i.
14             reverse(nums + n - k, nums + n);
15             
16             // Reverse all the numbers.
17             // Index i (0 <= i < n - k) becomes n - (n - k - i) = i + k.
18             // Index n - k + i (0 <= i < k) becomes n - (n - i) = i.
19             reverse(nums, nums + n);
20         }
21     };

 

解法三：

 1 public class Solution {
 2     public void rotate(int[] nums, int k) {
 3         k %= nums.length;
 4         reverse(nums, 0, nums.length - 1);
 5         reverse(nums, 0, k - 1);
 6         reverse(nums, k, nums.length - 1);
 7     }
 8     public void reverse(int[] nums, int start, int end) {
 9         while (start < end) {
10             int temp = nums[start];
11             nums[start] = nums[end];
12             nums[end] = temp;
13             start++;
14             end--;
15         }
16     }
17 }

先整体搞，再分开搞