题目
Given an array, rotate the array to the right by k
steps, where k
is non-negative.
Example 1:
Input: nums = [1,2,3,4,5,6,7], 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:
Input: nums = [-1,-100,3,99], 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]
Constraints:
1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105
Follow up:
- Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
- Could you do it in-place with
O(1)
extra space?
思路
如例子1的,
[1,2,3,4,5,6,7] => [5,6,7,1,2,3,4]
实际就是将最后的3个数字5,6,7和前面的1,2,3,4做了交换;
等价于
[1,2,3,4,5,6,7] => [7,6,5,4,3,2,1] 1)整体翻转 将前3个(就是当前的k值)和后面的分别单独翻转 [7,6,5] => [5, 6, 7] [4, 3, 2, 1] => [1, 2, 3, 4]
而数组翻转,只要两个指针从头尾开始,交换后,分别向中间移动,直到交叉或者会合。
代码
class Solution {
public void rotate(int[] nums, int k) {
int offset = k%nums.length;
if (offset == 0) return;
reverse(nums, 0, nums.length -1);
reverse(nums, 0, offset-1);
reverse(nums, offset, nums.length -1);
}
private void reverse(int[] nums, int left, int right) {
while(left < right) {
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
}
}
Be First to Comment