[LeetCode] 2348. Number of Zero-Filled Subarrays

Given an integer array nums, return the number of subarrays filled with 0.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,3,0,0,2,0,0,4]
Output: 6
Explanation: 
There are 4 occurrences of [0] as a subarray.
There are 2 occurrences of [0,0] as a subarray.
There is no occurrence of a subarray with a size more than 2 filled with 0. Therefore, we return 6.

Example 2:

Input: nums = [0,0,0,2,0,0]
Output: 9
Explanation:
There are 5 occurrences of [0] as a subarray.
There are 3 occurrences of [0,0] as a subarray.
There is 1 occurrence of [0,0,0] as a subarray.
There is no occurrence of a subarray with a size more than 3 filled with 0. Therefore, we return 9.

Example 3:

Input: nums = [2,10,2019]
Output: 0
Explanation: There is no subarray filled with 0. Therefore, we return 0.

Constraints:

• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109

全 0 子数组的数目。

给你一个整数数组 nums ，返回全部为 0 的 子数组 数目。

子数组 是一个数组中一段连续非空元素组成的序列。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/number-of-zero-filled-subarrays
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这是一道数学题，也可以用动态规划做。我参考了这个帖子。

具体的思路是如果我们当前位置上遇到的是一个 0，我们就把当前这个 0 当做子数组的结尾，来统计以当前这个 0 为结尾的符合题意的子数组有多少。举个例子，比如 [0, 0, 0, 0]，我们设一个变量 count 记录当前遇到的连续的 0 的个数

当我们遇到第一个 0 的时候，count = 1, res += count

当我们遇到第二个 0 的时候，count = 2, res += count

这里 count 其实暗含了子数组的个数。

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public long zeroFilledSubarray(int[] nums) {
 3         long res = 0;
 4         int n = nums.length;
 5         int count = 0;
 6         for (int i = 0; i < n; i++) {
 7             if (nums[i] == 0) {
 8                 count++;
 9                 res += count;
10             } else {
11                 count = 0;
12             }
13         }
14         return res;
15     }
16 }

LeetCode 题目总结