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.[show hint]
Hint:Could you 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 class Solution { 3 public: 4 void rotate(vector<int>& nums, int k) { //runtime:24ms 5 int n=nums.size(); 6 k %= n; 7 reverse(nums.begin(),nums.begin()+n-k); 8 reverse(nums.begin()+n-k,nums.end()); 9 reverse(nums.begin(),nums.end()); 10 } 11 };
Solution 2：超时
1 class Solution { 2 public: 3 void rotate(vector<int>& nums, int k) { 4 int n=nums.size(); 5 k %= n; 6 while(k--){ 7 int temp=nums[n-1]; 8 for(int i=n-1;i>0;i--){ 9 nums[i]=nums[i-1]; 10 } 11 nums[0]=temp; 12 } 13 } 14 };
Solution 3:
1 void rotate(int nums[], int n, int k) { 2 k = k % n; 3 if (k == 0) return; 4 int *temp = new int[n]; 5 memcpy(temp, nums+(n-k), sizeof(int)*k); 6 memcpy(temp+k, nums, sizeof(int)*(n-k)); 7 memcpy(nums, temp, sizeof(int)*n); 8 delete[] temp; 9 }