"Leetcode" 189-rotate Array

Rotate an array of n elements to the right by K steps.

For example, with n = 7 and k = 3, the array is [1,2,3,4,5,6,7] rotated to [5,6,7,1,2,3,4] .

Note:
Try to come up as many solutions as can, there is at least 3 different ways to solve this problem. [Show hint]

Hint:could do it in-place with O (1) extra space?

Related Problem:reverse Words in a String II

Solution 1:reverse[1,2,3,4]=[4,3,2,1], reverse[5,6,7]=[7,6,5], reverse[4,3,2,1,7,6,5]=[5,6,7,1,2,3,4]

1#include <algorithm>2 classSolution {3  Public:4     voidRotate (vector<int>& Nums,intK) {//runtime:24ms5         intn=nums.size ();6K%=N;7Reverse (Nums.begin (), Nums.begin () +n-k);8Reverse (Nums.begin () +n-k,nums.end ());9 Reverse (Nums.begin (), Nums.end ());Ten     } One};

Solution 2: Timeout

1 classSolution {2  Public:3     voidRotate (vector<int>& Nums,intk) {4         intn=nums.size ();5K%=N;6          while(k--){7             inttemp=nums[n-1];8              for(inti=n-1;i>0; i--){9nums[i]=nums[i-1];Ten             } Onenums[0]=temp; A         } -     } -};

Solution 3:

1 voidRotateintNums[],intNintk) {2K = k%N; 3     if(k = =0)return; 4     int*temp =New int[n]; 5memcpy (temp, nums+ (n-k),sizeof(int)*k); 6memcpy (Temp+k, Nums,sizeof(int) * (nk)); 7memcpy (Nums, temp,sizeof(int)*N); 8     Delete[] temp; 9}

"Leetcode" 189-rotate Array