Leetcode: insert Interval

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].

 1 /** 2  * Definition for an interval. 3  * public class Interval { 4  *     int start; 5  *     int end; 6  *     Interval() { start = 0; end = 0; } 7  *     Interval(int s, int e) { start = s; end = e; } 8  * } 9  */10 public class Solution {11     public List<Interval> insert(List<Interval> intervals, Interval newInterval) {12         if (newInterval == null) return intervals;13         if (intervals == null || intervals.size() == 0) {14             intervals.add(newInterval);15             return intervals;16         }17         int newleft = newInterval.start;18         int newright = newInterval.end;19         if (newleft > newright) return intervals;20         ArrayList<Interval> res = new ArrayList<Interval>();21         int lastleft = -1;22         int lastright = -1;23         int i = 0;24         for (; i<intervals.size(); i++) {25             Interval cur = intervals.get(i);26             if (newright < cur.start) {27                 if (lastleft == -1 && lastright == -1) {28                     lastleft = newleft;29                     lastright = newright;30                 }31                 else {32                     lastright = newright;33                 }34                 res.add(new Interval(lastleft, lastright));35                 res.add(cur);36                 break;37             }38             if (newleft < cur.start && newright >= cur.start && newright <= cur.end) {39                 if (lastleft == -1 && lastright == -1) {40                     lastleft = newleft;41                 }42                 lastright = cur.end;43                 res.add(new Interval(lastleft, lastright));44                 break;45             }46             if (newleft < cur.start && newright > cur.end) {47                 if (lastleft == -1 && lastright == -1) {48                     lastleft = newleft;49                 }50             }51             if (newleft >= cur.start && newright <= cur.end) {52                 res.add(cur);53                 break;54             }55             if (newleft >= cur.start && newleft <= cur.end && newright > cur.end) {56                 lastleft = cur.start;57             }58             if (newleft > cur.end) {59                 res.add(cur);60             }61         }62         for (i=i+1; i<intervals.size(); i++) {63             res.add(intervals.get(i));64         }65         if (newright > intervals.get(intervals.size()-1).end) {66             if (lastleft == -1 && lastright == -1) {67                 lastleft = newleft;68             }69             lastright = newright;70             res.add(new Interval(lastleft, lastright));71         }72         return res;73     }74 }

A simple practice of others:

 1 public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { 2     ArrayList<Interval> res = new ArrayList<Interval>(); 3     if(intervals.size()==0) 4     { 5         res.add(newInterval); 6         return res; 7     } 8     int i=0; 9     while(i<intervals.size() && intervals.get(i).end<newInterval.start)10     {11         res.add(intervals.get(i));12         i++;13     }14     if(i<intervals.size())15         newInterval.start = Math.min(newInterval.start, intervals.get(i).start);16     res.add(newInterval);17     while(i<intervals.size() && intervals.get(i).start<=newInterval.end)18     {19         newInterval.end = Math.max(newInterval.end, intervals.get(i).end);20         i++;21     }22     while(i<intervals.size())23     {24         res.add(intervals.get(i));25         i++;26     }27     return res;28 }

 

Leetcode: insert Interval