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
}


⬅️上一页

下一页➡️

Calendar Sep 6, 2020
Edit Edit this page
本站总访问量:  次 您是本站第  位访问者