[LeetCode] 189. Rotate Array 旋转数组
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?
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) 推荐解法2.
解法3：每次把最后一个元素移到第一位，后面的元素后移一位，循环往复，直到第k次。
解法4：交换最后k个元素和最开始的k个元素，在把前面的n-k个元素翻转。
解法参考：LeetCode Discuss
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);
}
};
C++: Swap the last k elements with the first k elements. Time complexity: O(n). Space complexity: O(1).
class Solution
{
public:
void rotate(int nums[], int n, int k)
{
for (; k = k%n; n -= k, nums += k)
{
// Swap the last k elements with the first k elements.
// The last k elements will be in the correct positions
// but we need to rotate the remaining (n - k) elements
// to the right by k steps.
for (int i = 0; i < k; i++)
{
swap(nums[i], nums[n - k + i]);
}
}
}
};
C++：Start from one element and keep rotating until we have rotated n different elements. Time complexity: O(n). Space complexity: O(1).
class Solution
{
public:
void rotate(int nums[], int n, int k)
{
if ((n == 0) || (k <= 0))
{
return;
}
int cntRotated = 0;
int start = 0;
int curr = 0;
int numToBeRotated = nums[0];
int tmp = 0;
// Keep rotating the elements until we have rotated n
// different elements.
while (cntRotated < n)
{
do
{
tmp = nums[(curr + k)%n];
nums[(curr+k)%n] = numToBeRotated;
numToBeRotated = tmp;
curr = (curr + k)%n;
cntRotated++;
} while (curr != start);
// Stop rotating the elements when we finish one cycle,
// i.e., we return to start.
// Move to next element to start a new cycle.
start++;
curr = start;
numToBeRotated = nums[curr];
}
}
};
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