<p>Given a binary array <code>nums</code>, return <em>the maximum number of consecutive </em><code>1</code><em>&#39;s in the array</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,0,1,1,0,1]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
      vector <int> s;
      int count =0;
      int m=0;
      for(int x=0;x<nums.size();x++){
     
        if(nums[x]==1){
        count+=1;
        }
        else  {
            s.push_back(count);
            count =0;
        }
        s.push_back(count);
      } 
        m = *max_element(s.begin(),s.end());
      return m; 
    }
};
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