Rotate Array

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.

<pre name="code" class="cpp" style="color: rgb(51, 51, 51); font-size: 24.44444465637207px; line-height: 25.98958396911621px;">class Solution {public:    void rotate(vector<int>& nums, int k) {      int temp;      for(int i=0;i<k;i++)      {          temp=nums[nums.size()-1];          for(int j=nums.size()-1;i>0;i--)          {              nums[i]=nums[i-1];          }          nums[0]=temp;      }    }};

如果是k步，那么就是把后面k个放到前面了嘛。

我们先把整个数组reverse，然后把前面的reverse回来，再把后面的reverse回来

对于AB我们要通过reverse操作得到BA

那么先把AB reverse一次得到reverse(B)reverse(A)

然后再把reverse(B),reverse(A)分别reverse一次就得到了BA

 

class Solution {public:    void rotate(vector<int>& nums, int k) {      k=k%(nums.size());       int n=nums.size();      int a[n];      for(int i=0;i<nums.size();i++)      {          a[i]=nums[i];      }             reverse(a, a + n);        reverse(a, a + k);        reverse(a + k, a + n);     for(int i=0;i<n;i++)      {          nums[i]=a[i];      }    }};

class Solution{public:void swap(int * a,int * b){int temp = *a;*a = *b;*b = temp;}void rotate(vector<int> & nums,int k){int length = nums.size();int l = length - (k % length);int r = k % length;for(int i = 0,j = l - 1;i < j;i++,j--)swap(&nums[i],&nums[j]);for(int i = l,j = length - 1;i < j;i++,j--)swap(&nums[i],&nums[j]);for(int i = 0,j = length - 1;i < j;i++,j--)swap(&nums[i],&nums[j]);}};