# [Insert Interval][title] ## Description 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]. ``` **Tags:** Array, Sort ## 思路 题意是给你一组有序区间,和一个待插入区间,让你待插入区间插入到前面的区间中,我们分三步走: 1. 首先把有序区间中小于待插入区间的部分加入到结果中; 2. 其次是插入待插入区间,如果有交集的话取两者交集的端点值; 3. 最后把有序区间中大于待插入区间的部分加入到结果中; ```java /** * 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; } * } */ class Solution { public List insert(List intervals, Interval newInterval) { if (intervals.isEmpty()) return Collections.singletonList(newInterval); List ans = new ArrayList<>(); int i = 0, len = intervals.size(); for (; i < len; ++i) { Interval interval = intervals.get(i); if (interval.end < newInterval.start) ans.add(interval); else break; } for (; i < len; ++i) { Interval interval = intervals.get(i); if (interval.start <= newInterval.end) { newInterval.start = Math.min(newInterval.start, interval.start); newInterval.end = Math.max(newInterval.end, interval.end); } else break; } ans.add(newInterval); for (; i < len; ++i) { ans.add(intervals.get(i)); } return ans; } } ``` ## 结语 如果你同我一样热爱数据结构、算法、LeetCode,可以关注我 GitHub 上的 LeetCode 题解:[awesome-java-leetcode][ajl] [title]: https://leetcode.com/problems/insert-interval [ajl]: https://github.com/Blankj/awesome-java-leetcode