[LeetCode] 189. Rotate Array 旋转数组

Posted 2020-10-28 轻风舞动

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?

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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