[Leetcode] 189. Rotate Array Rotation

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 [5,6,7,1,2,3,4] to.

Note:
Try to come up as many solutions as can, there is at least 3 different ways to solve this problem.

Hint:
Could 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.

Solution 1: Use an extra copy space. Time Complexity:o (n). Space complexity:o (N)

Solution 2: Flip the front n-k element, flip the remaining k elements, and finally flip all the elements. 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 & Lt End:            Nums[start], nums[end-1] = nums[end-1], Nums[start]            start + = 1            End-= 1

C + +: Make a 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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[Leetcode] 189. Rotate Array Rotation