Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [code][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]

[/code]

Example 2:

Input: [code][-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] [/code]

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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