Merge Intervals 合并区间@LeetCode

采用的是和Insert Interval一样的思想，只不过最开头要先排序一下，用到了java的Collections.sort(List<Interval> list, Comparator<? super Interval> c)

自己实现了一个Comparator的compare方法

http://blog.csdn.net/fightforyourdream/article/details/16876485

package Level4;import java.util.ArrayList;import java.util.Collections;import java.util.Comparator;import Utility.Interval;/** * 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].Discuss * */public class S56 {public static void main(String[] args) {ArrayList<Interval> intervals = new ArrayList<Interval>();intervals.add(new Interval(2,3));intervals.add(new Interval(3,4));intervals.add(new Interval(6,7));intervals.add(new Interval(1,10));System.out.println(merge(intervals));}public static ArrayList<Interval> merge(ArrayList<Interval> intervals) {ArrayList<Interval> ret = new ArrayList<Interval>();if(intervals.size() == 0){return ret;}// 要先按照start的顺序排序一下，这样就能保证全部能merge了！Collections.sort(intervals, new Comparator<Interval>() {@Overridepublic int compare(Interval o1, Interval o2) {return o1.start - o2.start;// 升序排}});Interval ref = intervals.get(0);for(int i=1; i<intervals.size(); i++){Interval cur = intervals.get(i);if(cur.end < ref.start){ret.add(cur);}else if(ref.end < cur.start){ret.add(ref);ref = cur;}else{ref = new Interval(Math.min(ref.start, cur.start), Math.max(ref.end, cur.end));}}ret.add(ref);return ret;}}

Second try:

import java.util.ArrayList;import java.util.Collections;import java.util.Comparator;/* 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]. */public class MergeIntervals {public static void main(String[] args) {}public ArrayList<Interval> merge(ArrayList<Interval> intervals) {if(intervals==null || intervals.size()==0){return intervals;}Collections.sort(intervals, new Comparator<Interval>() {@Overridepublic int compare(Interval o1, Interval o2) {return o1.start-o2.start;// 按照start的升序排列}});Interval merged = intervals.get(0);ArrayList<Interval> ret = new ArrayList<Interval>();for(int i=1; i<intervals.size(); i++){Interval cur = intervals.get(i);if(cur.end < merged.start){ret.add(cur);}else if(merged.end < cur.start){ret.add(merged);merged = cur;}else{int min = Math.min(cur.start, merged.start);int max = Math.max(cur.end, merged.end);merged = new Interval(min, max);}}ret.add(merged);return ret;}public static class Interval{int start;int end;public Interval(){start = 0;end = 0;}public Interval(int s, int e){start = s;end = e;}}}

/** * 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) {        List<Interval> ret = new ArrayList<Interval>();        if(intervals.size() == 0) {            return ret;        }                Collections.sort(intervals, new Comparator<Interval>(){            public int compare(Interval i1,Interval i2) {                return i1.start - i2.start;            }        } );                Interval merged = new Interval(intervals.get(0).start, intervals.get(0).end);        for(int i=1; i<intervals.size(); i++) {            Interval cur = intervals.get(i);            if(cur.start <= merged.end) {                merged = new Interval(Math.min(merged.start, cur.start), Math.max(merged.end, cur.end));            } else {                ret.add(merged);                merged = cur;            }        }        ret.add(merged);                return ret;    }}