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
java
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
java
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