Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Hint:
Could you do it in-place with O(1) extra space?
Could you do it in-place with O(1) extra space?
Related problem: Reverse Words in a String II
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
解法1：用一个额外的复制空间。 Time complexity: O(n). Space complexity: O(n)
解法2：翻转前n - k元素，翻转剩下的k个元素，最后翻转全部元素。O(n). Space complexity: O(1)
Java:
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
}
public void reverse(int[] nums, int start, int end) {
while (start < end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
}
Python:
class Solution:
def rotate(self, nums, k):
k %= len(nums)
self.reverse(nums, 0, len(nums))
self.reverse(nums, 0, k)
self.reverse(nums, k, len(nums))
def reverse(self, nums, start, end):
while start < end:
nums[start], nums[end - 1] = nums[end - 1], nums[start]
start += 1
end -= 1
C++: Make an extra copy and then rotate. Time complexity: O(n). Space complexity: O(n).
class Solution
{
public:
void rotate(int nums[], int n, int k)
{
if ((n == 0) || (k <= 0))
{
return;
}
// Make a copy of nums
vector<int> numsCopy(n);
for (int i = 0; i < n; i++)
{
numsCopy[i] = nums[i];
}
// Rotate the elements.
for (int i = 0; i < n; i++)
{
nums[(i + k)%n] = numsCopy[i];
}
}
};
C++: Reverse the first n - k elements, the last k elements, and then all the n elements.
class Solution
{
public:
void rotate(int nums[], int n, int k)
{
k = k%n;
// Reverse the first n - k numbers.
// Index i (0 <= i < n - k) becomes n - k - i.
reverse(nums, nums + n - k);
// Reverse tha last k numbers.
// Index n - k + i (0 <= i < k) becomes n - i.
reverse(nums + n - k, nums + n);
// Reverse all the numbers.
// Index i (0 <= i < n - k) becomes n - (n - k - i) = i + k.
// Index n - k + i (0 <= i < k) becomes n - (n - i) = i.
reverse(nums, nums + n);
}
};
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