标签：def span else ati with Plan sum == als
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].
题意：
给定一堆区间，插入一个新区间。
Solution1：Sort
code
1 /** 2 * Definition for an interval. 3 * public class Interval { 4 * int start; 5 * int end; 6 * Interval() { start = 0; end = 0; } 7 * Interval(int s, int e) { start = s; end = e; } 8 * } 9 */ 10 class Solution { 11 public List<Interval> insert(List<Interval> intervals, Interval newInterval) { 12 List<Interval> result = new ArrayList<Interval>(); 13 for (Interval i : intervals) { 14 if (newInterval == null || i.end < newInterval.start) 15 result.add(i); 16 else if (i.start > newInterval.end) { 17 result.add(newInterval); 18 result.add(i); 19 newInterval = null; 20 } else { 21 newInterval.start = Math.min(newInterval.start, i.start); 22 newInterval.end = Math.max(newInterval.end, i.end); 23 } 24 } 25 if (newInterval != null) 26 result.add(newInterval); 27 return result; 28 } 29 }
改版了新的signature之后：
[leetcode]57. Insert Interval插入区间
标签：def span else ati with Plan sum == als
原文地址：https://www.cnblogs.com/liuliu5151/p/10860172.html
