Insert Interval
问题描述
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]
.
解决思路
分三部分:
1. 把之前的较小区间加入list;
2. 更新重叠的区间的start和end,并加入新的区间;
3. 将剩余的区间加入。
程序
public class Solution { public List<Interval> insert(List<Interval> intervals, Interval newInterval) { if (newInterval == null) { return intervals; } List<Interval> merge = new ArrayList<Interval>(); if (intervals.size() == 0) { merge.add(newInterval); return merge; } int i = 0, n = intervals.size(); // add smaller while (i < n && intervals.get(i).end < newInterval.start) { merge.add(intervals.get(i++)); } // merge while (i < n && intervals.get(i).start <= newInterval.end) { newInterval.start = Math.min(newInterval.start, intervals.get(i).start); newInterval.end = Math.max(newInterval.end, intervals.get(i).end); ++i; } merge.add(newInterval); // add last while (i < n) { merge.add(intervals.get(i++)); } return merge; } }
Follow up
Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18]
,
return [1,6],[8,10],[15,18]
.
解决思路
1. 排序,从小到大;
2. merge。
程序
public class Solution { public List<Interval> merge(List<Interval> intervals) { List<Interval> mergeList = new ArrayList<Interval>(); if (intervals == null || intervals.isEmpty()) { return mergeList; } sortedIntervalsList(intervals); mergeList.add(intervals.get(0)); for (int i = 1; i < intervals.size(); i++) { Interval last = mergeList.get(mergeList.size() - 1); Interval insert = intervals.get(i); if (insert.start <= last.end) { last.end = Math.max(last.end, insert.end); } else { mergeList.add(insert); } } return mergeList; } private void sortedIntervalsList(List<Interval> intervals) { Comparator<Interval> comp = new Comparator<Interval>() { @Override public int compare(Interval i1, Interval i2) { if (i1.start > i2.start) { return 1; } else if (i1.start < i2.start) { return -1; } if (i1.end > i2.end) { return 1; } else if (i1.end < i2.end) { return -1; } return 0; } }; Collections.sort(intervals, comp); } }