3. Rotate the array
Given an array, move the elements in the array k positions to the right, where k is a non-negative number.
Method 1: own stupid way
class Solution {
public void rotate(int[] nums, int k) {
if(k>=0){
int length = nums.length;
k = k%length;
int num=0;
int[] nums1 = new int[length];
for(int i=0;i<length-k;i++){
nums1[i+k]=nums[i];
}
for(int j=length-k;j<length;j++){
nums1[num]=nums[j];
num++;
}
for(int m=0;m<length;m++){
nums[m] = nums1[m];
}
}
}
}
Method two: self-improved method
class Solution {
public void rotate(int[] nums, int k) {
if(k>=0){
int length=nums.length;
k = k%length;
int[] temp = new int[length];
for(int i=0;i<length;i++){
temp[(i+k)%length] = nums[i];
}
for(int i=0;i<length;i++){
nums[i] = temp[i];
}
}
}
}
Third, the official method 1-use additional arrays (similar to my method two)
class Solution {
public void rotate(int[] nums, int k) {
int n = nums.length;
int[] newArr = new int[n];
for (int i = 0; i < n; ++i) {
newArr[(i + k) % n] = nums[i];
}
System.arraycopy(newArr, 0, nums, 0, n);
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/rotate-array/solution/xuan-zhuan-shu-zu-by-leetcode-solution-nipk/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Method 4: Official Method 2-Array Flip
The space complexity of this method is O(1)
class Solution {
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
}
public void reverse(int[] nums, int start, int end) {
while (start < end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start += 1;
end -= 1;
}
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/rotate-array/solution/xuan-zhuan-shu-zu-by-leetcode-solution-nipk/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。