**Explanation**

For this question, we keep track of two variables: *max*, *count*. As we iterate through the elements of *nums*, *count* will keep track of the number of consecutive 1s. When *count* becomes greater than *max*, we update our *max* to the value of *count*.

We reset *count* back to 0 when we encounter a 0, and increment by 1 when we encounter a 1.

**Time Complexity**

The time complexity is O(n) where n is the number of elements in *nums*, since we iterate through all elements in *nums*.

```
class Solution {
public int findMaxConsecutiveOnes(int[] nums) {
int max = 0;
int count = 0;
// Iterate through all contents of num, updating count, max
for (int num : nums) {
if (num == 1) {
count += 1;
} else {
count = 0;
}
max = Math.max(max, count);
}
return max;
}
}
```