数组：合并区间

    给定若干个区间，对重叠的区间进行合并。如[1,3],[2,6],[8,10],[15,18]，合并后的区间为[1,6],[8,10],[15,18]。

    方法一：先把区间根据下界进行排序，然后判断重叠和合并

/**
 * 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) {
        Collections.sort(intervals, (Interval a, Interval b) -> (a.start - b.start));
        int i = 1;
        while(i < intervals.size()){
            if(intervals.get(i).start <= intervals.get(i - 1).end){
                intervals.get(i - 1).end = Math.max(intervals.get(i - 1).end, intervals.get(i).end);
                intervals.remove(i);
            }
            else i++;
        }
        return intervals;
    }
}

    方法二：把区间转化为两个一维数组，分别排序，根据数组的大小来判断重叠。因为直接对一维数组进行排序，效率更高。

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;
}