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?
Solutions
My solution (also optimal)
Idea: directly put numbers in their new positions.
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n = nums.size();
if(n == 0 || k % n == 0) return;
k %= n;
int count = 0; // number of elements in correct position
int start = 0;
while(count < n){
int i = start;
int prev = nums[i]; // value of number in previous index
do{
i = (i + k) % n; // destination index
swap(prev, nums[i]);
++count;
}while(i != start);
++start;
}
}
};
Easier to understand solution
Idea: the objective is to move the first n - k numbers to the last n - k positions, and move the last k numbers to the first k positions. If we reverse nums, then the original first n - k numbers and the last k numbers are in correct ranges. Then we reverse the two ranges to make all numbers in correct positions.
E.g.,
Original array: [1, 2, 3, 4, 5, 6] , k = 4
Objective: [3, 4, 5, 6, 1, 2]
After first reversing: [6, 5, 4, 3, 2, 1].[6, 5, 4, 3] and [2, 1] are in correct ranges but in reversed order.
class Solution {
public:
void rotate(vector<int>& nums, int k) {
int n = nums.size();
if(n == 0 || k % n == 0) return;
k %= n;
reverse(nums.begin(), nums.end());
reverse(nums.begin(), nums.begin() + k);
reverse(nums.begin() + k, nums.end());
}
};
