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].
解题思路一：
用两个指针startIndex和endIndex来维护每次添加intervals的start和end的位置，然后分类讨论即可，JAVA实现如下：
public List<Interval> merge(List<Interval> intervals) {
List<Interval> list = new ArrayList<Interval>();
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.start > 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;
}
解题思路二：
首先构造一个比较器，对interval按照start进行排序，然后进行遍历，在遍历过程中，如果结果集合为空或者当前interval与结果集合中的最后一个interval不重叠，那么就直接将当前interval直接加入到结果集合中；如果发生了重叠，那么将结果集合的最后一个interval的右端点改为当前interval的右端点，JAVA实现如下：
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)
list.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;
}