[leedcode 56] Insert Interval

標籤：

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].

/** * 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) {        //遍曆intervals，並和newInterval對比：        //1 如果比newInterval小：直接加入新結果集合        //2 如果有overlap，動態改變newInterval為新的區間，繼續合并        //3 如果比newInterval大：加入newInterval到新集合，然後把newInterval更新為當前對象        //解題思路：畫圖，先排除兩種不重疊的情況，對重疊情況最後處理，        //需要注意中間值temp的變化，以及最後需要添加最後一個temp        List<Interval> res=new ArrayList<Interval>();        Interval temp=newInterval;        for(int i=0;i<intervals.size();i++){                        Interval cur=intervals.get(i);            if(cur.start>temp.end){                res.add(temp);                temp=cur;            }else{                if(temp.start>cur.end){                    res.add(cur);                }else{                    int start=Math.min(cur.start,temp.start);                    int end=Math.max(cur.end,temp.end);                    Interval newInt=new Interval(start,end);                    temp=newInt;                }            }                                }        res.add(temp);        return res;    }}

[leedcode 56] Insert Interval