package demo_1118_MergeIntervals;
/**
* Given a collection of intervals,merge all overlaping intervals.
* For example,
* Given [1,3],[2,6],[8,10],[15,18]
* return [1,6],[8,10],[15,18]
*/
//解析:
import sun.swing.BakedArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.List;
/**
* 提议:
* 有很多的区间,把有重叠的却见的合并
* 思路:
* 先排序,然后检查响铃的2个区间的,看起那样给区间的结尾是哦㱎后一个区间的开始,注意前一个区间的b
* 包含的后一个区间的情况。
*
* 用Java的自带的sort ()方法,只要自己重写compare ()方法即可。
*/
/**
* 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 class MyComparator implements Comparator<Interval> {
@Override
public int compare(Interval a, Interval b) {
return a.start - b.start;
}
}
public List<Interval> merge(List<Interval> intervals) {
List<Interval> ans = new ArrayList<Interval>();
if (intervals.size() == 0) return ans;
Collections.sort(intervals, new MyComparator());
int start = intervals.get(0).start;
int end = intervals.get(0).end;
for (int i = 0; i < intervals.size(); i++) {
Interval inter = intervals.get(i);
if (inter.start > end) {
ans.add(new Interval(start, end));
start = inter.start;
end = inter.end;
} else {
end = Math.max(end, inter.end);
}
}
ans.add(new Interval(start, end));
return ans;
}
}
Solution_mergerInterViews
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转载自blog.csdn.net/xiamaocheng/article/details/84198084
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