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.

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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.

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Constraints:

• 1 <= nums.length <= 105

• -109?<= nums[i] <= 109

最終祭出了數(shù)學(xué)公式，，，當(dāng)有連續(xù)n個(gè)0的時(shí)候，判斷有多少個(gè)組合，就是n*(n+1)/2，然后累加，

因?yàn)樽詈笠粋€(gè)可能算不到，因?yàn)樽詈笠晃豢赡苓€是0，所以結(jié)束for循環(huán)的時(shí)候，還要加一下判斷。確認(rèn)此時(shí)cnt是否為0，不為0，還要加cnt*(cnt+1)/2;

最后返回結(jié)果即可。

Runtime:?6 ms, faster than?34.17%?of?Java?online submissions for?Number of Zero-Filled Subarrays.

Memory Usage:?60.2 MB, less than?20.69%?of?Java?online submissions for?Number of Zero-Filled Subarrays.