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Solution106.java
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83 lines (79 loc) · 2.75 KB
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/*
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].
*/
/*
class Interval {
int start;
int end;
Interval() {start = 0, end = 0;}
Interval(int s, int e) {start = s; end = e;}
}
*/
public class Solution {
public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
if (null == newInterval || newInterval.start > newInterval.end) return intervals;
if (intervals.size()==0) {
intervals.add(newInterval);
return intervals;
}
int cnt = 0;
for (Interval inter : intervals) {
if (newInterval.end < inter.start) break;
else {
if (newInterval.end <= inter.end) {
if (newInterval.start > inter.start) {
newInterval.start = inter.start;
}
newInterval.end = inter.end;
cnt++;
} else {
if (newInterval.start <= inter.end) {
if (newInterval.start >= inter.start) {
newInterval.start = inter.start;
}
cnt++;
}
}
}
}
if (cnt == 0) {
if (newInterval.start > intervals.get(intervals.size()-1).end) {
intervals.add(newInterval);
return intervals;
}
if (newInterval.end < intervals.get(0).start) {
intervals.add(0, newInterval);
return intervals;
}
int i = 0;
for (; i < intervals.size(); i++) {
if (newInterval.end < intervals.get(i).start) {
break;
}
}
intervals.add(i, newInterval);
return intervals;
} else {
ArrayList<Interval> rst = new ArrayList<Interval>();
int i = 0;
for (; i < intervals.size(); i++) {
if (newInterval.start > intervals.get(i).end) {
rst.add(intervals.get(i));
} else if (newInterval.end < intervals.get(i).start) {
break;
}
}
rst.add(newInterval);
for (; i < intervals.size(); i++) {
rst.add(intervals.get(i));
}
return rst;
}
}
}