chapter-four/0001~0099/0057.Insert-Interval

57. Insert Interval

题目

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].

题目大意

这一题是第 56 题的加强版。给出多个没有重叠的区间,然后再给一个区间,要求把如果有重叠的区间进行合并。

解题思路

可以分 3 段处理,先添加原来的区间,即在给的 newInterval 之前的区间。然后添加 newInterval ,注意这里可能需要合并多个区间。最后把原来剩下的部分添加到最终结果中即可。

代码

package leetcode /** * Definition for an interval. * type Interval struct { * Start int * End int * } */ func insert(intervals []Interval, newInterval Interval) []Interval { res := make([]Interval, 0) if len(intervals) == 0 { res = append(res, newInterval) return res } curIndex := 0 for curIndex < len(intervals) && intervals[curIndex].End < newInterval.Start { res = append(res, intervals[curIndex]) curIndex++ } for curIndex < len(intervals) && intervals[curIndex].Start <= newInterval.End { newInterval = Interval{Start: min(newInterval.Start, intervals[curIndex].Start), End: max(newInterval.End, intervals[curIndex].End)} curIndex++ } res = append(res, newInterval) for curIndex < len(intervals) { res = append(res, intervals[curIndex]) curIndex++ } return res }