LeetCode Rotate Array，leetcoderotate

LeetCode Rotate Array，leetcoderotate

Rotate Array Total Accepted: 12759 Total Submissions: 73112 My Submissions Question Solution
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.

題意：迴圈數組，n代表數組的長度，k代表向右移動的次數。
解法一：

class Solution {public:    void rotate(int nums[], int n, int k) {        if(n==0)return;        k=k%n;//當k大於n的時候，n次迴圈會回到初始位置，因此，可以省略若干次        if (k == 0) return;          int *s=new int[k];//為了一步到位的展開移動，申請k個額外空間用於儲存被移出去的元素        for(int i=0;i<k;++i)            s[i]=nums[n-k+i];//儲存被移出去的元素        for(int j=n-k-1;j>=0;--j)            nums[j+k]=nums[j];//移動        for(int i=0;i<k;++i)            nums[i]=s[i];//被移出的元素進行歸位        free(s);    }};

需要額外空間O（k%n）
33 / 33 test cases passed.
Status: Accepted
Runtime: 29 ms

解法二（網路擷取）：
三次翻轉法，第一次翻轉前n-k個，第二次翻轉後k個，第三次翻轉全部。

class Solution {public:    void rotate(int nums[], int n, int k) {        if(n==0)return ;        k=k%n;        if(k==0)return ;        reverse(nums,n-k,n-1);        reverse(nums,0,n-k-1);        reverse(nums,0,n-1);    }    void reverse(int nums[],int i,int j)    {        for(;i<j;++i,--j)        {            int t=nums[i];            nums[i]=nums[j];            nums[j]=t;        }    }};

33 / 33 test cases passed.
Status: Accepted
Runtime: 26 ms