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:
把每个以 nums[i]为终点的 subarray 的数量加起来， 假设 count[i]所有的以 nums[i]为终点的 subarray 的数量， 则 count[i+1] = count[i] + 1, 因为所有以 nums[i]为终点的 subarray 在加上 nums[i+1]这个元素之后都会转变成以 nums[i+1]为终点的 subarray， 那个+1 是指的 nums[i+1]本身也是一个 subarray。
impl Solution {
pub fn zero_filled_subarray(nums: Vec<i32>) -> i64 {
let mut ans = 0;
let mut count = 0;
for n in nums {
if n == 0 {
count += 1;
ans += count;
continue;
}
count = 0;
}
ans
}
}