[LeetCode] Rotate Array，leetcoderotate

[LeetCode] Rotate Array，leetcoderotate

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.

解題思路1

首先把數組複製一遍，然後找到元素之間的映射關係： newnum[i] = oldnum[(i - k + n) % n]，時間複雜度為O(n)，空間複雜度為O(n)。

實現代碼1

/*****************************************************************    *  @Author   : 楚興    *  @Date     : 2015/2/24 16:58    *  @Status   : Accepted    *  @Runtime  : 33 ms******************************************************************/class Solution {public:    void rotate(int nums[], int n, int k) {        int *temp = new int[n];        memcpy(temp, nums, n * sizeof(int));        k = k % n;        for (int i = 0; i < n; i++)        {            nums[i] = temp[(i - k + n) % n];        }        delete [] temp;    }};

解題思路2

將數組看成是一個環，每個元素每次往前走一步，迴圈k次。時間複雜度為O(k*n)，耗時較長，空間複雜度為O(1)。

實現代碼2

/*****************************************************************    *  @Author   : 楚興    *  @Date     : 2015/2/24 17:10    *  @Status   : Accepted    *  @Runtime  : 872 ms******************************************************************/class Solution {public:    void rotate(int nums[], int n, int k) {        k = k % n;        while (k--)        {            int temp = nums[n - 1];            for (int i = n - 1; i > 0; i--)            {                nums[i] = nums[i - 1];            }            nums[0] = temp;        }    }};

解題思路3

①將整個數組反轉
②將由分割點分割的兩個數組分別反轉
即：1 2 3 4 5 6 7 -> 7 6 5 | 4 3 2 1 -> 5 6 7 | 1 2 3 4
時間複雜度為O(n)，空間複雜度為O(1)。

實現代碼3

/*****************************************************************    *  @Author   : 楚興    *  @Date     : 2015/2/24 17:39    *  @Status   : Accepted    *  @Runtime  : 25 ms******************************************************************/class Solution {public:    void rotate(int nums[], int n, int k) {        k = k % n;        rev(nums, 0, n - 1);        rev(nums, 0, k - 1);        rev(nums, k, n - 1);    }    void rev(int num[], int left, int right)    {        int temp;        while (left < right)        {            temp = num[left];            num[left++] = num[right];            num[right--] = temp;        }    }};