Merge Intervals

Javascript (ES6)

/**
 * @param {number[][]} intervals
 * @return {number[][]}
 */
const merge = (intervals) => {
  intervals.sort((a, b) => a[0] - b[0]);
  const ret = [];
  let prev = null;
  for (const interval of intervals) {
    if (prev != null) {
      if (interval[0] > prev[1]) {
        ret.push(prev);
        prev = interval;
      } else {
        // Merge case.
        prev[1] = Math.max(interval[1], prev[1]);
      }
    } else {
      prev = interval;
    }
  }
  if (prev !== null) {
    ret.push(prev);
  }
  return ret;
};

console.log(merge([
  [1, 3],
  [2, 6],
  [8, 10],
  [15, 18]
])); // [[1,6],[8,10],[15,18]]

Demo

If  “n” is the length of the input list, then the “Time Complexity”  is O(nlogn).

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */

class Solution {
 public List < Interval > merge(List < Interval > intervals) {
  // Sort the input list by the start times.
  Collections.sort(intervals, (a, b) -> a.start - b.start);
  List < Interval > ret = new ArrayList < Interval > ();
  Interval prev = null;
  for (Interval curr: intervals) {
   if (prev == null) {
    prev = curr;
   } else {
    if (curr.start <= prev.end) {
     prev.end = Math.max(curr.end, prev.end);
    } else {
     ret.add(prev);
     prev = curr;
    }
   }
  }
  if (prev != null) {
   ret.add(prev);
  }
  return ret;
 }
}