57. Insert Interval
Medium
You are given an array of non-overlapping intervals
intervals
where intervals[i] = [starti, endi]
represent the start and the end of the ith
interval and intervals
is sorted in ascending order by starti
. You are also given an interval newInterval = [start, end]
that represents the start and end of another interval.Insert
newInterval
into intervals
such that intervals
is still sorted in ascending order by starti
and intervals
still does not have any overlapping intervals (merge overlapping intervals if necessary).Return
intervals
after the insertion.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].
Constraints:
0 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 105
intervals
is sorted bystarti
in ascending order.newInterval.length == 2
0 <= start <= end <= 105
Runtime: 10 ms, faster than 81.36%
Memory Usage: 4.6 MB, less than 76.10%
func insert(intervals [][]int, newInterval []int) [][]int {
l := len(intervals)
for i, interval := range intervals {
if interval[0] > newInterval[0] || (interval[0] == newInterval[0] && interval[1] > newInterval[1]) {
intervals = append(intervals[:i+1], intervals[i:]...)
intervals[i] = newInterval
break
}
}
if l==len(intervals) {
intervals = append(intervals, newInterval)
}
ans := make([][]int, 0)
ans = append(ans, intervals[0])
index := 0
for i:=1; i<len(intervals); i++ {
if intervals[i][1] <= ans[index][1] { continue }
if intervals[i][0] <= ans[index][1] {
ans[index][1] = intervals[i][1]
} else {
ans = append(ans, intervals[i])
index++
}
}
return ans
}
Last modified 6mo ago