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.
Solution 1: note that line 6 uses k%size. the time complexity of erase is O(n), so the worst total time could be O(n^2)
1 class Solution { 2 public: 3 void rotate(vector<int>& nums, int k) { 4 int size=nums.size(); 5 int i=0; 6 while (i<size-k%size){ 7 nums.push_back(nums[0]); 8 nums.erase(nums.begin()); 9 i++; 10 } 11 } 12 };
Solution 2:reverse the previous n-k nums and then reverse the later k nums. Finally reverse all. O(n)
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1 class Solution { 2 public: 3 void rotate(vector<int>& nums, int k) { 4 if (nums.empty() || (k %= nums.size()) == 0) return; 5 int n = nums.size(); 6 reverse(nums.begin(), nums.begin() + n - k); 7 reverse(nums.begin() + n - k, nums.end()); 8 reverse(nums.begin(), nums.end()); 9 } 10 };