3.4 KiB
Leetcode Rotate-Array
2022-07-10 08:54
Algorithms:
#algorithm #array_in_place_operation #reverse_array
Data structures:
#DS #array
Difficulty:
#coding_problem #difficulty-medium
Additional tags:
#leetcode #CS_list_need_practicing #CS_list_need_understanding
Revisions:
N/A
Related topics:
Links:
Problem
Given an array, rotate the array to the right by k
steps, where k
is non-negative.
Examples
Example 1:
Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]
Constraints
1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
0 <= k <= 105
Thoughts
Method 1: Array In-place operation
This one is hard. ==#TODO: Revisit this one some other day==
- Reversing a subarray doesn't change the continuity, the neighbors inside the subarray is same.
- Reversing a subarray changes the edge's neighbors
Suppose a array looks like:
[1, 2,| 3, 4, 5]
If reverse all two subarrays, the neighbors isn't changed only the orders are changed
[1, 2,| 3, 4, 5]: 5-1-2, 1-2-3, 2-3-4, 4-5-1
[2, 1,| 5, 4, 3]: 3-2-1, 2-1-5, 5-1-4, 4-3-2
When reversed back, the order is changed back to original, but the two sub-arrays are swapped
[3, 4, 5,| 1, 2]
By doing this, the order in inner-array isn't changed, but the order of all is changed.
Method 2: Copy
Watch out for integer overflow.
Method 3: CPP's STL sway subarrays
Remember to sanitize k, when k > size
Solution
Method 1:
class Solution {
void swap(vector<int> &nums, int l, int r) {
int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
void reverse(vector<int> &nums, int l, int r) {
while (l < r) {
swap(nums, l, r);
l++;
r--;
}
}
public:
void rotate(vector<int> &nums, int k) {
const int size = nums.size();
k = k % size;
// Reversion
reverse(nums, 0, size - k - 1);
reverse(nums, size - k, size - 1);
reverse(nums, 0, size - 1);
}
};
Method 2:
class Solution {
void swap(vector<int> &nums, int l, int r) {
int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
public:
void rotate(vector<int> &nums, int k) {
int size = nums.size();
if (size <= 1) {
return;
}
// Copy
k = size - k;
while (k < 0) {
k += size;
}
vector<int> ans(nums.size());
for (int i = 0; i < size; i++) {
ans[i] = nums[(i + k) % size];
}
for (int i = 0; i < size; i++) {
nums[i] = ans[i];
}
}
};
Method 3:
class Solution {
void swap(vector<int> &nums, int l, int r) {
int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
public:
void rotate(vector<int> &nums, int k) {
int size = nums.size();
if (size <= 1) {
return;
}
// STL
k = k % size;
k = size - k;
while (k < 0) {
k += size;
}
nums.insert(nums.end(), nums.begin(), nums.begin() + k);
nums.erase(nums.begin(), nums.begin() + k);
}
};