<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> myNums;
        bool ones = false;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == 1) {
                if (ones) {
                    myNums[myNums.size() - 1]++;
                }
                else {
                    myNums.push_back(1);
                    ones = true;
                }
            } else {
                ones = false;
            }
        }
        int max = 0;
        for (int i = 0; i < myNums.size(); i++) {
            if (myNums[i] > max) {max = myNums[i];}
        }
        return max;
    }
};
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