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].
此题主要考虑到各种边界条件
可以用一些小的例子累得出循环的条件，注意将删除节点后应将i--
public class Solution { public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) { if (intervals.isEmpty()) { intervals.add(newInterval); return intervals; } Interval pre = newInterval; for(int i=0;i<intervals.size();i++){ int start = pre.start; int end = pre.end; Interval temp = intervals.get(i); if (temp.start > end) { intervals.add(i, pre); return intervals; } else if (temp.start == end) { temp.start = start; return intervals; } else { if (start <= temp.start && end <= temp.end) { temp.start = start; return intervals; } else if (start <= temp.start && end > temp.end) { intervals.remove(temp); i--; continue; } else if (start > temp.start && end < temp.end) { return intervals; } else if (start > temp.start &&start<temp.end&& end >=temp.end) { pre.start = temp.start; pre.end = end; intervals.remove(temp); i--; continue; } else if (start > temp.end) { pre.start = start; pre.end = end; continue; } else if (start == temp.end) { pre.start = temp.start; pre.end = end; intervals.remove(temp); i--; continue; } } } intervals.add(pre); return intervals; } }
【LeetCode】Insert Interval,布布扣,bubuko.com
原文：http://www.cnblogs.com/yixianyixian/p/3721781.html