演算法插入間隔

dongyu2013發表於2014-05-04
演算法

Problem:

Given a set of non-overlapping & sorted intervals, insert a new interval into the intervals (merge if necessary). 

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].

Thoughts of This Problem

Quickly summarize 3 cases. Whenever there is intersection, created a new interval.

insert-interval

Java Solution 

/**
 * 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 ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {  
        ArrayList<Interval> result = new ArrayList<Interval>();  
 for(Interval interval: intervals)
{ 
 if(interval.end < newInterval.start){ result.add(interval); }
 else if(interval.start > newInterval.end){ result.add(newInterval); newInterval = interval; }
 else if(interval.end >= newInterval.start || interval.start <= newInterval.end)
 { newInterval = new Interval(Math.min(interval.start, newInterval.start), Math.max(newInterval.end, interval.end)); } }  
        result.add(newInterval);   
 return result; 
 } 
}

來自 “ ITPUB部落格 ” ，連結：http://blog.itpub.net/29012686/viewspace-1153201/，如需轉載，請註明出處，否則將追究法律責任。

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