标签：eval style last desc get order inline insert turn
Given a non-overlapping interval list which is sorted by start point.
Insert a new interval into it, make sure the list is still in order and non-overlapping (merge intervals if necessary).
Insert (2, 5) into [(1,2), (5,9)], we get [(1,9)].
Insert (3, 4) into [(1,2), (5,9)], we get [(1,2), (3,4), (5,9)].
题意：给定一个区间，将它插进一个有序的区间集合里，新的区间依然要保持有序性。这就需要考虑到区间的合并问题，我们可以定义一个新的集合，来存放最后的结果。定义一个temp游标，用一个循环从旧的集合中依次取出区间，与待插入区间进行比较。那么如何比较呢？假定新区间的end都小于temp的start，那说明新区间比temp要小，那么直接将新区间放进结果集合里就行了，剩下的依次插入。不然，则说明需要进行区间的合并，具体代码如下：
/**
* Definition of Interval:
* public classs Interval {
* int start, end;
* Interval(int start, int end) {
* this.start = start;
* this.end = end;
* }
* }
*/
public class Solution {
/**
* @param intervals: Sorted interval list.
* @param newInterval: new interval.
* @return: A new interval list.
*/
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
// write your code here
//特殊情况的讨论
List<Interval>ans=new ArrayList<Interval>();
if(intervals.size()==0){
ans.add(newInterval);
return ans;
}
if(newInterval==null){
return intervals;
}
if(newInterval.start>intervals.get(intervals.size()-1).end){
intervals.add(newInterval);
return intervals;
}
//一般情况的讨论
Interval last=null;
for(int i=0;i<intervals.size();i++){
//用不到newIneval
Interval temp=intervals.get(i);
if(newInterval.start>temp.end){
ans.add(temp);
continue;
}else{
//分两种情况
if(newInterval.end<temp.start){
ans.add(newInterval);
last=temp;
}else{
int start=newInterval.start<temp.start?newInterval.start:temp.start;
int end=newInterval.end<temp.end?temp.end:newInterval.end;
//合并
last=new Interval(start,end);
}
//对剩下的进行处理
for(int j=i+1;j<intervals.size();j++){
Interval t=intervals.get(j);
if(last.end<t.start){
//归并完成
ans.add(last);
last=t;
}else{
//继续归并
last.end=last.end>t.end?last.end:t.end;
}
}
ans.add(last);
break;
}
}
return ans;
}
}
30. Insert Interval【LintCode by java】
标签：eval style last desc get order inline insert turn
原文地址：https://www.cnblogs.com/phdeblog/p/9111079.html
