LeetCode每日一题(2090. K Radius Subarray Averages)

You are given a 0-indexed array nums of n integers, and an integer k.

The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k (inclusive). If there are less than k elements before or after the index i, then the k-radius average is -1.

Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i.

The average of x elements is the sum of the x elements divided by x, using integer division. The integer division truncates toward zero, which means losing its fractional part.

For example, the average of four elements 2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.

Example 1:

Input: nums = [7,4,3,9,1,8,5,2,6], k = 3
Output: [-1,-1,-1,5,4,4,-1,-1,-1]

Explanation:

• avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index.
• The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.
  Using integer division, avg[3] = 37 / 7 = 5.
• For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.
• For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.
• avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.

Example 2:

Input: nums = [100000], k = 0
Output: [100000]

Explanation:

• The sum of the subarray centered at index 0 with radius 0 is: 100000.
  avg[0] = 100000 / 1 = 100000.

Example 3:

Input: nums = [8], k = 100000
Output: [-1]

Explanation:

• avg[0] is -1 because there are less than k elements before and after index 0.

Constraints:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i], k <= 105

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维持一个队列, 将 nums 中的数字依次放入队列中， 队列中始终保持着 2 * k + 1 个元素， 这样每次入队出队，我们就可以计算出 avg[i-k-1], 其中 i 为指向 nums 的下一个入队元素的 index

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impl Solution {
    pub fn get_averages(nums: Vec<i32>, k: i32) -> Vec<i32> {
        let mut ans = vec![-1; nums.len()];
        if (nums.len() as i32) < 2 * k + 1 {
            return ans;
        }
        let mut i = 0;
        let mut stack = Vec::new();
        let mut sum = 0i64;
        while i < 2 * k as usize + 1 {
            let v = nums[i];
            stack.push(v);
            sum += v as i64;
            i += 1;
        }
        ans[i - k as usize - 1] = (sum / stack.len() as i64) as i32;
        while i < nums.len() {
            stack.push(nums[i]);
            sum += nums[i] as i64;
            let p = stack.remove(0);
            sum -= p as i64;
            i += 1;
            ans[i - k as usize - 1] = (sum / stack.len() as i64) as i32;
        }
        ans
    }
}
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