Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
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:
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?
// Using extra space
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(n)
vector<int> tmp = nums;
int n = nums.size();
for (int i = 0; i < n; ++i) {
nums[(i + k) % n] = tmp[i];
}
}
// Constant version of the previous method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
if (nums.empty() || (k %= nums.size()) == 0) return;
int start_idx = 0, idx = 0, pre = 0, cur = nums[0], n = nums.size();
for (int i = 0; i < n; ++i) {
pre = cur;
idx = (idx + k) % n;
cur = nums[idx];
nums[idx] = pre;
if (idx == start_idx) {
start_idx = ++idx;
cur = nums[start_idx];
}
}
}
// Reverse method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
if (nums.empty() || ((k %= nums.size()) == 0)) return;
int n = nums.size();
reverse(nums.begin(), nums.end());
reverse(nums.begin(), nums.begin() + k);
reverse(nums.begin() + k, nums.end());
}
// Reverse method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
if (nums.empty() || ((k %= nums.size()) == 0)) return;
int n = nums.size();
reverse(nums.begin(), nums.begin() + n - k);
reverse(nums.begin() + n - k, nums.end());
reverse(nums.begin(), nums.end());
}
// Swap numbers to correct positions and keep doing so for the next part of array
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
int n = nums.size(), start = 0;
while (n && (k %= n)) {
for (int i = 0; i < k; ++i) {
swap(nums[i + start], nums[n - k + i + start]);
}
n -= k;
start += k;
}
}