问题描述:
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]
.
原问题链接:https://leetcode.com/problems/merge-intervals/
问题分析
这里需要将一组interval给合并起来。其实如果有一组interval它们的起始点start是递增的,我们可以通过取第一个开始,每次和后面的比较,如果后面那个interval和前面的没有交叉,则将前面的这个interval加入到结果集中,否则进行合并。而合并的流程比较简单,取两个interval中间end值大的那个。因为这里是按照start的顺序从小到大排列的,所以甚至可以不需要比较。
这样,在具体的实现里我们首先将给定的List按照start从小到大的排序,然后再按照前面描述的合并以及加入到结果集中。具体的实现如下:
/** * 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> merge(List<Interval> intervals) { if(intervals.size() <= 1) return intervals; Collections.sort(intervals, new Comparator<Interval>() { public int compare(Interval t1, Interval t2) { return t1.start - t2.start; } }); List<Interval> result = new ArrayList<Interval>(); int curStart = intervals.get(0).start; int curEnd = intervals.get(0).end; for(int i = 1; i < intervals.size(); i++) { if(curEnd >= intervals.get(i).start) { curEnd = Math.max(curEnd, intervals.get(i).end); curStart = Math.min(curStart, intervals.get(i).start); } else { result.add(new Interval(curStart, curEnd)); curStart = intervals.get(i).start; curEnd = intervals.get(i).end; } } result.add(new Interval(curStart, curEnd)); return result; } }
因为目前的实现已经支持java 8了,如果我们再运用一点lambda表达式语法的技巧, 代码可以更加简略一点:
public class Solution { public List<Interval> merge(List<Interval> intervals) { if(intervals.size() <= 1) return intervals; Collections.sort(intervals, (t1, t2) -> t1.start - t2.start); List<Interval> result = new ArrayList<Interval>(); int curStart = intervals.get(0).start; int curEnd = intervals.get(0).end; for(int i = 1; i < intervals.size(); i++) { if(curEnd >= intervals.get(i).start) { curEnd = Math.max(curEnd, intervals.get(i).end); curStart = Math.min(curStart, intervals.get(i).start); } else { result.add(new Interval(curStart, curEnd)); curStart = intervals.get(i).start; curEnd = intervals.get(i).end; } } result.add(new Interval(curStart, curEnd)); return result; } }