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].

Tips:

注意要先按照interval的start排序，要重写compareTo方法。

先把第一个加进去，然后判断下一个interval的开始如果小于前一个的end的话，就不加这次的interval，改result里面的end为当前interval的end和result已存最后一个interval的end的最大值。

Code:

/**
 * 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; }
 * }
 */
public class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        List<Interval> result = new ArrayList<>();
        if (intervals == null || intervals.size() < 2) {
            return intervals;
        }
        Collections.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval a, Interval b) {
                return a.start - b.start;
            } 
        });
        for (int i = 0; i < intervals.size(); i++) {
            if (result.size() == 0) {
                result.add(intervals.get(i)); 
                continue;
            } 
            if (intervals.get(i).start <= result.get(result.size() - 1).end) {
                result.get(result.size() - 1).end = Math.max(result.get(result.size() - 1).end, intervals.get(i).end);
            } else {
                result.add(intervals.get(i));
            }
        }
        return result;
    }
}