LeetCode-----57. Insert Interval(数组插入并重新合并)

来源：互联网发布：手机相贯线软件编辑：程序博客网时间：2024/06/01 07:25

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:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

Subscribe to see which companies asked this question

/** * Definition for an interval. * public class Interval { *     int start; *     int end; *     Interval() { start = 0; end = 0; } *     Interval(int s, int e) { start = s; end = e; } * } */public class Solution {    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {        if (newInterval == null || intervals == null) {            return intervals;        }        List<Interval> results = new ArrayList<Interval>();        int insertPos = 0;        for (Interval interval : intervals) {            if (interval.end < newInterval.start) {                results.add(interval);                insertPos++;            } else if (interval.start > newInterval.end) {                results.add(interval);            } else {                newInterval.start = Math.min(interval.start, newInterval.start);                newInterval.end = Math.max(interval.end, newInterval.end);            }        }        results.add(insertPos, newInterval);        return results;    }}

0 0