[leetcode]57. Insert Interval(Java)

https://leetcode.com/problems/insert-interval/#/description

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]

.

Java Code:

package go.jacob.day627;

import java.util.ArrayList;
import java.util.List;

public class Demo1 {
/*
* Runtime: 15 ms.Your runtime beats 77.22 % of java submissions.
*/
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
//res存取最终结果
List<Interval> res = new ArrayList<Interval>();
int index = 0;
//当intervals中interval区间与newInterval没有重叠时，直接添加到res集合中
while (index < intervals.size() && newInterval.start > intervals.get(index).end)
res.add(intervals.get(index++));
//当存在重叠时，将newInterval赋予新的对象
while (index < intervals.size() && newInterval.end >= intervals.get(index).start) {
newInterval=new Interval(
Math.min(newInterval.start, intervals.get(index).start),
Math.max(newInterval.end, intervals.get(index).end));
index++;
}
res.add(newInterval);
//如果intervals中还有interval，直接添加到res中
while (index < intervals.size())
res.add(intervals.get(index++));

return res;
}
}

class Interval {
int start;
int end;

Interval() {
start = 0;
end = 0;
}

Interval(int s, int e) {
start = s;
end = e;
}
}