Leetcode 56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

1.普通sort方法：

public List<Interval> merge(List<Interval> intervals) {    // sort start&end    int n = intervals.size();    int[] starts = new int[n];    int[] ends = new int[n];    for (int i = 0; i < n; i++) {        starts[i] = intervals.get(i).start;        ends[i] = intervals.get(i).end;    }    Arrays.sort(starts);    Arrays.sort(ends);    // loop through    List<Interval> res = new ArrayList<Interval>();    for (int i = 0, j = 0; i < n; i++) { // j is start of interval.        if (i == n - 1 || starts[i + 1] > ends[i]){            res.add(new Interval(starts[j], ends[i]));            j = i + 1;        }    }    return res;}

2.use collections comparator in java 8

public List<Interval> merge(List<Interval> intervals) {        if (intervals.size() <= 1)        return intervals;        // Sort by ascending starting point using an anonymous Comparator        intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start));        List<Interval> result = new LinkedList<Interval>();        int start = intervals.get(0).start;        int end = intervals.get(0).end;        for (Interval interval : intervals) {            if (interval.start <= end) // Overlapping intervals, move the end if needed                end = Math.max(end, interval.end);            else {                     // Disjoint intervals, add the previous one and reset bounds                result.add(new Interval(start, end));                start = interval.start;                end = interval.end;            }        }        // Add the last interval        result.add(new Interval(start, end));        return result;    }