问题描述 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].
code
/**
* 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; }
* }
*/
/*
https://leetcodenotes.wordpress.com/2013/10/19/insert-interval/
思路清晰比什么都重要
先插入，再合并删除
*/
public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
int index = binaryByStart(intervals, newInterval);
intervals.add(index, newInterval);
Iterator<Interval> it = intervals.iterator();
Interval pre = it.next();
while(it.hasNext()) {
Interval now = it.next();
boolean remove = false;
if(within(pre,now)) remove = true;
else if(overlap(pre,now)) {
pre.end = now.end;
remove = true;
}
if(remove) it.remove();
else pre = now;
}
return intervals;
}
public int binaryByStart(List<Interval> intervals, Interval newInterval) {
int lo = 0,hi = intervals.size()-1;
int mid = 0;
while(lo <= hi) {
mid = (hi+lo)/2;
if(intervals.get(mid).start <= newInterval.start) {
lo = mid+1;
continue;
} else if(intervals.get(mid).start > newInterval.start) {
hi = mid-1;
continue;
}
}
return lo;
}
private boolean overlap(Interval a,Interval b) {
return a.end >= b.start;
}
private boolean within(Interval a,Interval b) {
return a.end >= b.end;
}
}