Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

 ``````1 2 3 4 5 6 `````` ``````Input: [1,2,3,4,5,6,7] and k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5] rotate 3 steps to the right: [5,6,7,1,2,3,4]``````

Example 2:

 ``````1 2 3 4 5 `````` ``````Input: [-1,-100,3,99] and k = 2 Output: [3,99,-1,-100] Explanation: rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100]``````

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

eg: 1,2,3,4,5,6,7 k=3

1. 反转前半部分

4,3,2,1,5,6,7

1. 反转后半部分

4,3,2,1,7,6,5

1. 反转整个数组

5,6,7,1,2,3,4

Solution 1

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 `````` ``````pub fn rotate(nums: &mut Vec, k: i32) { if nums.is_empty() || k <= 0 { return; } let o_len = nums.len(); let mod_k = k as usize % o_len; reverse(nums, 0, o_len - mod_k - 1); reverse(nums, o_len - mod_k, o_len - 1); reverse(nums, 0, o_len - 1); } pub fn reverse(nums: &mut Vec, start: usize, end: usize) { let mut o_start = start; let mut o_end = end; while o_start < o_end { nums.swap(o_start, o_end); o_start += 1; o_end -= 1; } } ``````

Solution 2

api 解法,效率不高,但好看

 `````` 1 2 3 4 5 6 7 8 9 10 `````` ``````pub fn rotate(nums: &mut Vec, k: i32) { if nums.is_empty() { return; } let mod_k = k % nums.len() as i32; for _ in 0..mod_k as usize { let item = nums.pop().unwrap(); nums.insert(0, item); } } ``````