![https://leetcode.com/problems/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].
Example 3:
Input: intervals = [], newInterval = [5,7]
Output: [[5,7]]
Example 4:
Input: intervals = [[1,5]], newInterval = [2,3]
Output: [[1,5]]
Example 5:
Input: intervals = [[1,5]], newInterval = [2,7]
Output: [[1,7]]
Constraints:
- 0 <= intervals.length <= 104
- intervals[i].length == 2
- 0 <= intervals[i][0] <= intervals[i][1] <= 105
- intervals is sorted by intervals[i][0] in ascending order.
- newInterval.length == 2
- 0 <= newInterval[0] <= newInterval[1] <= 105
class Solution {
public int[][] insert(int[][] intervals, int[] newInterval) {
List<int[]> res = new ArrayList<>();
int index = 0;
int n = intervals.length;
while (index < n && !overlaps(intervals[index], newInterval) && intervals[index][1] < newInterval[0]) {
res.add(intervals[index]);
index++;
}
int overlapIndex = index;
while (overlapIndex < n && overlaps(intervals[overlapIndex], newInterval)) {
newInterval = merge(intervals[overlapIndex], newInterval);
overlapIndex++;
}
res.add(index, newInterval);
index = overlapIndex;
while (index < n && !overlaps(intervals[index], newInterval)) {
res.add(intervals[index]);
index++;
}
return res.toArray(new int[res.size()][]);
}
int[] merge(int[] a, int[] b) {
return new int[] {Math.min(a[0], b[0]), Math.max(a[1], b[1])};
}
boolean overlaps(int[] a, int[] b) {
return Math.max(a[0], b[0]) <= Math.min(a[1],b[1]);
}
}