189. Rotate Array【easy】

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：

 class Solution {
 public:
     void rotate(vector<int>& nums, int k) {
         int len = nums.size();

         if (len ==  || k % len == )
         {
             return;
         }

         int mid = len - k % len;

         reverseArray(nums, , mid - );
         reverseArray(nums, mid, len - );
         reverseArray(nums, , len - );
     }

     void reverseArray(vector<int>& nums, int start, int end)
     {
         int s = start;
         int e = end;

         while (s <= e)
         {
             int temp = nums[s];
             nums[s] = nums[e];
             nums[e] = temp;
             s++;
             e--;
         }
     }
 };

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：

 class Solution
     {
     public:
         void rotate(int nums[], int n, int k)
         {
             k = k%n;

             // Reverse the first n - k numbers.
             // Index i (0 <= i < n - k) becomes n - k - i.
             reverse(nums, nums + n - k);

             // Reverse tha last k numbers.
             // Index n - k + i (0 <= i < k) becomes n - i.
             reverse(nums + n - k, nums + n);

             // Reverse all the numbers.
             // Index i (0 <= i < n - k) becomes n - (n - k - i) = i + k.
             // Index n - k + i (0 <= i < k) becomes n - (n - i) = i.
             reverse(nums, nums + n);
         }
     };

解法三：

 public class Solution {
     public void rotate(int[] nums, int k) {
         k %= nums.length;
         reverse(nums, , nums.length - );
         reverse(nums, , k - );
         reverse(nums, k, nums.length - );
     }
     public void reverse(int[] nums, int start, int end) {
         while (start < end) {
             int temp = nums[start];
             nums[start] = nums[end];
             nums[end] = temp;
             start++;
             end--;
         }
     }
 }

先整体搞，再分开搞