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?
Subject to the effect :
Given an array, the array "right" k bits.
Understanding:
An array [6, 7] and k = 3 as an example, the result is shifted right three [5,6,7,1,2,3,4].
The array is divided into two parts, left to nk-1 0 is a part, nk n to another part.
Retrograde two parts, respectively the elements, the result is [4,3,2,1,7,6,5]; then reverse the entire array is set, the result is [5,6,7,1,2,3,4].
Code C ++:
class Solution { public: void rotate(vector<int>& nums, int k) { int n = nums.size(); if(k>=n) k = k%n; reverse(nums.begin(),nums.begin()+n-k); reverse(nums.begin()+n-k,nums.end()); reverse(nums.begin(),nums.end()); } };
operation result:
When execution: 44 ms, beat the 40.35% of all users to submit in C ++
Memory consumption: 9.6 MB, defeated 14.06% of all users to submit in C ++