# 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 by `starti` in ascending order.
• `newInterval.length == 2`
• `0 <= start <= end <= 105`

### 解題

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 > newInterval || (interval == newInterval && interval > newInterval) {
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)
index := 0
for i:=1; i<len(intervals); i++ {
if intervals[i] <= ans[index] { continue }
if intervals[i] <= ans[index] {
ans[index] = intervals[i]
} else {
ans = append(ans, intervals[i])
index++
}
}
return ans
}