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
解題
Runtime: 10 ms, faster than 81.36%
Memory Usage: 4.6 MB, less than 76.10%
第二次解
func insert(intervals [][]int, newInterval []int) [][]int {
if len(intervals) == 0 { // intervals 中沒東西的情況
return [][]int{ newInterval }
}
// 將新給的 interval 加入 intervals 陣列後重新進行排序
intervals = append(intervals, newInterval)
sort.Slice(intervals, func(i, j int) bool {
if intervals[i][0] == intervals[j][0] {
return intervals[i][1] < intervals[j][1]
}
return intervals[i][0] < intervals[j][0]
})
// stack
ans := make([][]int, 0)
ans = append(ans, intervals[0])
// 每次插入新的 element 時做檢查
for i:=1; i<len(intervals); i++ {
last := ans[len(ans)-1]
current := intervals[i]
if current[0] > last[1] {
ans = append(ans, current)
} else { //current[0] <= last[1]
if current[1] > last[1] {
last[1] = current[1]
}
}
}
return ans
}
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