Merge Intervals - Implementation

Here is the implementation:

function merge(intervals):
    sort intervals by start
    result = [intervals[0]]

    for i from 1 to intervals.length - 1:
        last = result[result.length - 1]
        current = intervals[i]

        if current[0] <= last[1]:
            last[1] = max(last[1], current[1])
        else:
            result.append(current)

    return result

Sorting takes $O(n \log n)$. The merge pass is $O(n)$. Total: $O(n \log n)$ time, $O(n)$ space for output.