力扣（leetcode）每日一题 3255 长度为 K 的子数组的能量值 II|滑动窗口

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 复制代码

  
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 复制代码

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