<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) {
        stack<int> st;
        int maxCount = 0;
        if(nums.size() <= 0) return maxCount; 
        maxCount = nums[0] == 1 ? 1 : 0;
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] == 0) {
                st.push(maxCount);
                maxCount = 0;
                continue;
            }
            maxCount++;
        }
        while(!st.empty()){
            auto max = st.top();
            st.pop();
            if(max > maxCount){
                maxCount = max;
            }
        }
        return maxCount;
    }
};
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