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?
//使用双端队列 想怎么干怎么干 和利用的空间比较多
public static int[] rotate(int[] nums, int k) {
// ArrayDeque, ConcurrentLinkedDeque, LinkedBlockingDeque, LinkedList
Deque deque = new ArrayDeque();
for (int num:nums){
deque.add(num);
}
for (int j = 0;j<k;j++){
int last = (int) deque.getLast();
deque.removeLast();
deque.addFirst(last);
}
int m = 0;
while (deque.peek()!= null){
nums[m] = (int) deque.pop();
m++;
}
System.out.println(Arrays.toString(nums));
return nums;
}
public static int[] rotate2(int[] nums, int k) {
k = k % nums.length;
int[] kHolder = new int[k];
System.arraycopy(nums, nums.length - k, kHolder, 0, k);
System.arraycopy(nums, 0, nums, k, nums.length - k);
System.arraycopy(kHolder, 0, nums, 0, k);
return kHolder;
}
public static int[] rotate3(int[] nums, int k) {
if(k==0 || nums.length <2) {
return nums;
}
int len = nums.length;
int r = k % len;
int[] temp = new int[len];
for(int j=0; j<len; j++){
temp[(j+r)%len] = nums[j];
}
System.arraycopy(temp, 0, nums, 0, temp.length);
return temp;
}
题目地址：https://leetcode.com/problems/rotate-array/