【leetcode】56. Merge Intervals

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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].

/** * 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; } * } */   //思路：先排序，然后检查相邻两个区间，看前一个区间的结尾是否大于后一个区间的开始，注意前一个区间包含后一个区间的情况。//用Java自带的sort()方法，只要自己重写compare()方法即可。public class Solution {    public List<Interval> merge(List<Interval> intervals) {        if (intervals == null || intervals.size() <= 1){            return intervals;        }        Collections.sort(intervals,new IntervalComparator());                ArrayList<Interval> result = new ArrayList<Interval>();        Interval last = intervals.get(0);        for (int i = 1; i < intervals.size(); i++){            Interval curt = intervals.get(i);            if (curt.start <= last.end){                last.end = Math.max(last.end,curt.end);            } else {                result.add(last);                last = curt;            }        }        result.add(last);        return result;    }            private class IntervalComparator implements Comparator<Interval>{        public int compare(Interval a, Interval b){              return a.start - b.start;        }    }}

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