Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
Example 1:
Input: intervals = [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: intervals = [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Constraints:
1 <= intervals.length <= 104intervals[i].length == 20 <= starti <= endi <= 104
Idea:
- First step (Important): sort the intervals by the first number
- Use result array to store the first interval
- Compare the last interval of the result array vs. new coming interval
- new interval > last interval of the result array : store the new interval into result array
- new interval < last interval of the result array : store the interval that has longer range
Solution:
/**
* @param {number[][]} intervals
* @return {number[][]}
*/
var merge = function(intervals) {
intervals.sort(([a, b], [c, d]) => a - c);
let res = [];
for (let interval of intervals) {
if (res.length === 0 || interval[0] > res[res.length - 1][1]) {
res.push(interval.concat());
} else {
res[res.length - 1][1] = Math.max(interval[1], res[res.length - 1][1]);
}
}
return res;
};