3255. Find the Power of K-Size Subarrays II
1.题干
You are given an array of integers nums of length n and a positive integer k.
The power of an array is defined as:
- Its maximum element if all of its elements are consecutive and sorted in ascending order.
- -1 otherwise.
You need to find the power of all
subarrays
of nums of size k.
Return an integer array results of size n - k + 1, where results[i] is the power of nums[i..(i + k - 1)].
Example 1:
Input: nums = [1,2,3,4,3,2,5], k = 3
Output: [3,4,-1,-1,-1]
Explanation:
There are 5 subarrays of nums of size 3:
[1, 2, 3]with the maximum element 3.[2, 3, 4]with the maximum element 4.[3, 4, 3]whose elements are not consecutive.[4, 3, 2]whose elements are not sorted.[3, 2, 5]whose elements are not consecutive.
Example 2:
Input: nums = [2,2,2,2,2], k = 4
Output: [-1,-1]
Example 3:
Input: nums = [3,2,3,2,3,2], k = 2
Output: [-1,3,-1,3,-1]
2.题解
滑动窗口，常规解法
public static int[] resultsArray(int[] nums, int k) { //3 LinkedList<Integer> list = new LinkedList<>(); for (int i = 0; i < k - 1; i++) { // 保证队列一次从小到大差值1排列 while (!list.isEmpty() && nums[list.getLast()] != (nums[i] - 1)) { list.pollLast(); } list.add(i); } int[] res = new int[nums.length - k + 1]; for (int i = k - 1; i < nums.length; i++) { // 保证队列一次从小到大差值1排列 while (!list.isEmpty() && nums[list.getLast()] != (nums[i] - 1)) { list.pollLast(); } list.add(i); // 如果队列数量和k相等，就是成立的 if (list.size() == k) { res[i - k + 1] = nums[i]; } else { res[i - k + 1] = -1; } // 加入当前为index 6，然后队列长度k为3，那么对于index 4的数据进行弹出，确保下个循环index 7进来之后，队列长度维持在k为3 if (list.getFirst() == i - k + 1) { list.pollFirst(); } } return res;
}
官方题解，属于技巧性
public int[] resultsArray(int[] nums, int k) { int n = nums.length; int[] ans = new int[n - k + 1]; Arrays.fill(ans, -1); int cnt = 0; for (int i = 0; i < n; i++) { cnt = i == 0 || nums[i] - nums[i - 1] != 1 ? 1 : cnt + 1; // 当连续数的时候，累计加上1 if (cnt >= k) { ans[i - k + 1] = nums[i]; } } return ans;
}
3.总结
面试编程时候肯定思维紧张，往滑动窗口上靠就对了
