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Dorchester Center, MA 02124

Given a binary array `nums`

, return *the maximum number of consecutive *`1`

*‘s in the array*.

**Example 1:**

```
Input: nums = [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.
```

**Example 2:**

```
Input: nums = [1,0,1,1,0,1]
Output: 2
```

**Constraints:**

`1 <= nums.length <= 10`

^{5}`nums[i]`

is either`0`

or`1`

.

```
public int findMaxConsecutiveOnes(int[] nums) {
int maxHere = 0, max = 0;
for (int n : nums)
max = Math.max(max, maxHere = n == 0 ? 0 : maxHere + 1);
return max;
}
```

```
class Solution(object):
def findMaxConsecutiveOnes(self, nums):
cnt = 0
ans = 0
for num in nums:
if num == 1:
cnt += 1
ans = max(ans, cnt)
else:
cnt = 0
return ans
```

```
int findMaxConsecutiveOnes(vector<int>& nums) {
int max_cnt = 0, cnt = 0;
for (auto n : nums) {
if (n == 1) max_cnt = max(++cnt, max_cnt);
else cnt = 0;
}
return max_cnt;
}
```

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