57-Insert Interval

【题目】

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

【analyze】

1.利用之前类似的一道题目的做法

2.先根据区间的start对区间进行排序

3.再依次插入

【算法】

/**
 * 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> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> res=new ArrayList<Interval>();
        intervals.add(newInterval);
        Collections.sort(intervals,new Comparator<Interval>() {
            public int compare(Interval i1,Interval i2) {
                return i1.start-i2.start;
            }
        });
        Interval pre=intervals.get(0);
        for(int i=1;i<intervals.size();i++) {
            Interval cur=intervals.get(i);
            if(pre.end>=cur.start) {
                Interval merge=new Interval(pre.start,Math.max(pre.end,cur.end));
                pre=merge;
            }else {
                res.add(pre);
                pre=cur;
            }
        }
        res.add(pre);
        return res;
    }
}