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 * 104
-231 <= nums[i] <= 231 - 1
0 <= k <= 105
Related Topics:
Array
Similar Questions:
// OJ: https://leetcode.com/problems/rotate-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
class Solution {
public:
void rotate(vector<int>& A, int k) {
int N = A.size();
k %= N;
if (k == 0) return;
vector<int> tmp(k);
for (int i = 0; i < k; ++i) tmp[i] = A[N + i - k];
for (int i = N - k - 1; i >= 0; --i) A[i + k] = A[i];
for (int i = 0; i < k; ++i) A[i] = tmp[i];
}
};
// OJ: https://leetcode.com/problems/rotate-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
void rotate(vector<int>& A, int k) {
int cnt = 0, N = A.size();
k %= N;
if (k == 0) return;
for (int i = 0; cnt < N; ++i) {
int j = i, tmp = A[j];
do {
int t = (j + k) % N, next = A[t];
A[t] = tmp;
tmp = next;
j = t;
++cnt;
} while (j != i);
}
}
};
// OJ: https://leetcode.com/problems/rotate-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
void rotate(vector<int>& A, int k) {
k = (2 * A.size() - k) % A.size();
if (k == 0) return;
reverse(begin(A), begin(A) + k);
reverse(begin(A) + k, end(A));
reverse(begin(A), end(A));
}
};