Description
Given an array, rotate the array to the right by k steps, where k is non-negative.
Follow up:
- 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?
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 <= 2 * 10^4- It's guaranteed that
nums[i]fits in a 32 bit-signed integer. k >= 0
Analyse
将数组nums的后k位翻转到数组起始位置
Input: [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4] // 5, 6, 7翻转到数组首部
申请临时空间
原数组与翻转后的数组下标有对应关系
i -> (i + k) % nums.length
public void rotate(int[] nums, int k) {
int len = nums.length;
k = k % len;
int[] tmp = new int[len];
for (int i = 0; i < len; ++i) {
tmp[(i + k) % len] = nums[i];
}
for (int j = 0; j < len; ++j) {
nums[j] = tmp[j];
}
}
reverse三次
k = 3
1 2 3 4 5 6 7 // expected 5 6 7 1 2 3 4
4 3 2 1 5 6 7 // reverse 1 2 3 4
4 3 2 1 7 6 5 // reverse 5 6 7
5 6 7 1 2 3 4 // reverse 全部
另一种三次reverse
k = 3
1 2 3 4 5 6 7 // expected 5 6 7 1 2 3 4
4 3 2 1 5 6 7 // reverse 1 2 3 4
7 6 5 1 2 3 4 // reverse 全部
5 6 7 1 2 3 4 // reverse 5 6 7
public void reverse(int[] nums, int low, int high) {
while (low < high) {
int tmp = nums[low];
nums[low] = nums[high];
nums[high] = tmp;
++low;
--high;
}
}
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - k - 1);
reverse(nums, nums.length - k, nums.length-1);
reverse(nums, 0, nums.length-1);
}