Problem
Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5] rotate 3 steps to the right: [5,6,7,1,2,3,4]
Example 2:
Input:[-1,-100,3,99] and k = 2 Output: [3,99,-1,-100] Explanation: rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100]
Note:
- Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
- Could you do it in-place with O(1) extra space?
Solution 1: Simulate rotation with three reverses.
If k >= n, rotating k times has the same effect as rotating k % n times.
[1,2,3,4,5,6,7], K = 3
[5,6,7,1,2,3,4]
We can simulate the rotation with three reverses.
- reverse the whole array O(n) [7,6,5,4,3,2,1]
- reverse the left part 0 ~ k – 1 O(k) [5,6,7,4,3,2,1]
- reverse the right part k ~ n – 1 O(n-k) [5,6,7,1,2,3,4]
Time complexity: O(n)
Space complexity: O(1) in-place
C++
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class Solution { public: void rotate(vector<int>& nums, int k) { if (nums.empty()) return; k %= nums.size(); if (k == 0) return; reverse(begin(nums), end(nums)); reverse(begin(nums), begin(nums) + k); reverse(begin(nums) + k, end(nums)); } }; |
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