【LeetCode刷题记录】Rotate Array

题目：Rotate Array 

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.

解答：

题意是实现数组循环右移k位，这题主要是考察空间复杂度。

解法一：Copy法

最直接的方法就是开辟一个temp[n]数组，暂存nums[n]元素，方便移位赋值，但这种方法空间复杂度为O(n)。最开始，我是分[0,n-k-1]和[n-k,n-1]两部分处理，如下所示：

void rotate(int nums[], int n, int k) {    int temp[n], i;    k = k%n; // k modulo n to limit k in [0,n-1]    for(i=0; i<n; i++)        temp[i] = nums[i];    for(i=0; i<n-k; i++)        nums[i+k] = temp[i];    for(i=n-k; i<n; i++)        nums[i-n+k] = temp[i];}

在review讨论区的解答时，发现这两部可以合并成一部分，只需要引入取模操作。优化代码如下所示：

void rotate(int nums[], int n, int k) {    if ((n == 0) || (k <= 0)) // exception handle is essential        return;    int temp[n], i;    k = k%n; // k modulo n to limit k in [0,n-1]    for(i=0; i<n; i++)        temp[i] = nums[i];    for(i=0; i<n; i++)        nums[(i+k)%k] = temp[i];}

解法二：递推移位法

单独考虑每一次移位操作，假设a[7] = {1,2,3,4,5,6,7}，k = 3。那么，a[0]要移位到a[3]，即a[3] = a[0]。同时，原来的a[3]要移位到a[6]，所以在进行a[0]移位前，必须用一个中间数temp暂存a[3]的值。以此类推，一直到要移位的元素下标回到最初的地方(a[0])。这种方法空间复杂度为O(1)参考的讨论区代码如下：

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];            }        }

更多解法参考链接：

Summary of C++ solutions