[leetcode] 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].

https://oj.leetcode.com/problems/merge-intervals/

思路：先按照起始点排序，然后遍历合并即可。

代码：

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

public class Solution {
public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
if (intervals == null)
return null;
Collections.sort(intervals, new Comparator<Interval>() {

public int compare(Interval a, Interval b) {
return a.start - b.start;
}

});
ArrayList<Interval> res = new ArrayList<Interval>();

int i;
for (i = 0; i < intervals.size();) {
int j = i + 1;

Interval merged = new Interval(intervals.get(i).start,
intervals.get(i).end);
while (j < intervals.size()
&& isConnected(merged, intervals.get(j))) {
merged = merge(merged, intervals.get(j));
j++;
}
res.add(merged);
i = j;
}

return res;
}

private boolean isConnected(Interval a, Interval b) {
return (a.start >= b.start && a.start <= b.end)
|| (b.start >= a.start && b.start <= a.end);

}

private Interval merge(Interval a, Interval b) {
int newStart = a.start < b.start ? a.start : b.start;
int newEnd = a.end > b.end ? a.end : b.end;
a.start = newStart;
a.end = newEnd;
return a;
}

public static void main(String[] args) {
// [1,3],[2,6],[8,10],[15,18]
Interval one = new Interval(2, 4);
Interval two = new Interval(1, 2);
Interval three = new Interval(4, 10);
Interval four = new Interval(10, 18);
ArrayList<Interval> intervals = new ArrayList<Interval>();
intervals.add(one);
intervals.add(two);
intervals.add(three);
intervals.add(four);

System.out.println(new Solution().merge(intervals));
}

}

参考：

http://leetcodenotes.wordpress.com/2013/08/01/merge-intervals/