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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
Note
找到插入的起点:直到条件不成立intervals.get(i).end < newInterval.start
插入:只要intervals.get(i).start <= newInterval.end
就更新start和end
newInterval.start = Math.min(intervals.get(i).start, newInterval.start);
newInterval.end = Math.max(intervals.get(i).end, newInterval.end);
加入剩下的区间
Code
class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
if (newInterval == null) {
return intervals;
}
List<Interval> res = new ArrayList<>();
int i = 0;
int len = intervals.size();
while (i < len && intervals.get(i).end < newInterval.start) {
res.add(intervals.get(i++));
}
while (i < len && intervals.get(i).start <= newInterval.end) {
newInterval.start = Math.min(intervals.get(i).start, newInterval.start);
newInterval.end = Math.max(intervals.get(i).end, newInterval.end);
i++;
}
res.add(newInterval);
while (i < len) {
res.add(intervals.get(i++));
}
return res;
}
}
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