LeetCode Rotate Array「建议收藏」

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.

题意：循环数组，n代表数组的长度，k代表向右移动的次数。
解法一：

class Solution {
public:
    void rotate(int nums[], int n, int k) {
        if(n==0)return;
        k=k%n;//当k大于n的时候。n次循环会回到初始位置，因此，能够省略若干次
        if (k == 0) return;  
        int *s=new int[k];//为了一步到位的展开移动，申请k个额外空间用于保存被移出去的元素
        for(int i=0;i<k;++i)
            s[i]=nums[n-k+i];//保存被移出去的元素
        for(int j=n-k-1;j>=0;--j)
            nums[j+k]=nums[j];//移动
        for(int i=0;i<k;++i)
            nums[i]=s[i];//被移出的元素进行归位
        free(s);
    }
};

须要额外空间O（k%n）
33 / 33 test cases passed.
Status: Accepted
Runtime: 29 ms

解法二（网络获取）：
三次翻转法，第一次翻转前n-k个。第二次翻转后k个，第三次翻转所有。

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