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.

題目給我們一組數列，目標是返回所有由0組成的子數列的數量！

我的解題思路：

• 子數列是數列中連續的數

1. 重頭開始遍歷數列
2. 看到0再開始紀錄有連續幾個0
3. 因為題目沒有說要排除重複的組合
4. 所以每個0的數列都要計算
5. 遍歷完數列後返回0子數列的數量

• 題目範例 :
  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.

class Solution {
    public long zeroFilledSubarray(int[] nums) {
        int result = 0;
        int count = 0;

        // 開始遍歷
        for (int num : nums) {
            if (num == 0) {
                count++;
                result += count;
            } else{
                // 如果數字不是 0，
                count =0;
            }
        }
        return result;
    }
}

因為之前已經寫過幾次跟計算子數列有關的題目，加上題目跟之前比起來也算簡單～
所以這次有經驗就寫比較順！話說不知不覺鐵人賽已經開始三週了，時間過真快。