Java for Leetcode 056 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] .

Problem solving idea One:

Using two pointers startindex and endindex to maintain the position of the start and end of each add intervals, and then categorize the discussion, the Java implementation is as follows:

Public list<interval> Merge (list<interval> intervals) {list<interval> List = new Arraylist<in Terval> (); if (intervals.size () = = 0) return List;list.add (Intervals.get (0)); for (int i = 1; i < intervals.size (); i++ ) {Interval temp = intervals.get (i); int startIndex = 0, EndIndex = 0;for (int j = 0; J < List.size (), j + +) {if (temp.st Art > List.get (j). End) {StartIndex + = 2;endindex + 2;continue;} if (Temp.end < List.get (j). Start) break;if (Temp.start >= List.get (j). Start) startindex++;if (Temp.end > List.get (j). End) {EndIndex + = 2;continue;} if (Temp.end >= list.get (j). Start) Endindex++;break;} if (startindex==endindex&&startindex%2==0) List.add (startindex/2,new Interval (temp.start,temp.end)); else if (startindex%2==0&&endindex%2==0) {list.get (STARTINDEX/2). Start=temp.start;list.get (STARTINDEX/2). end= temp.end;for (int k=1;k<endindex/2-startindex/2;k++) list.remove (startindex/2+1);} else if (startindex%2==0&&endindex%2!=0) {LIST.GET (STARTINDEX/2). Start=temp.start;list.get (STARTINDEX/2). End=list.get (ENDINDEX/2). end;for (int k=1;k<= endindex/2-startindex/2;k++) List.remove (startindex/2+1);} else if (startindex%2!=0&&endindex%2==0) {list.get (STARTINDEX/2). end=temp.end;for (int k=1;k<endindex/2- startindex/2;k++) List.remove (startindex/2+1);} else if (startindex%2!=0&&endindex%2!=0) {list.get (STARTINDEX/2). End=list.get (ENDINDEX/2). end;for (int k=1; k<=endindex/2-startindex/2;k++) List.remove (startindex/2+1);}} return list;}

Two ways to solve problems:

First, a comparator is constructed, the interval is sorted by start, and then traversed, and if the result set is empty or interval does not overlap with the last one in the result set, the current is directly interval interval added to the result set. If an overlap occurs, the last right endpoint of the result collection is changed to the interval current interval right endpoint, and the Java implementation is as follows:

Public list<interval> Merge (list<interval> intervals) {        list<interval> List = new arraylist< Interval> ();        comparator<interval> Comparator = new comparator<interval> () {            @Override public            int Compare ( Interval O1, Interval O2) {                if (O1.start = = O2.start)                    return o1.end-o2.end;                return o1.start-o2.start;            }        ;        Collections.sort (intervals, comparator);        for (Interval interval:intervals)            if (list.size () = = 0 | | list.get (list.size ()-1). End < Interval.start)                 L Ist.add (New Interval (Interval.start, Interval.end));             else             List.get (list.size ()-1). End = Math.max (Interval.end, List.get (List.size ()-1). end);        return list;    }

Java for Leetcode 056 Merge intervals