435. Non-overlapping Intervals

最新推荐文章于 2024-02-19 15:34:13 发布

原创最新推荐文章于 2024-02-19 15:34:13 发布 · 304 阅读

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本文介绍了一种通过移除最少数量的区间使剩余区间不重叠的方法。提供了具体实例及算法实现，采用贪心策略，先按区间的起始点排序，再逐个判断并保留右端点值较小的区间。

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Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

You may assume the interval's end point is always bigger than its start point.
Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]

Output: 1

Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

Input: [ [1,2], [1,2], [1,2] ]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

贪心法则，先把区间按照起点为第一关键字，终点为第二关键字排序。

排序以后挨个区间来考虑，当前为第K个区间，如果第K+1个区间和第K个区间有重叠，我们需要保留一个区间，

应该保留的区间是右端点值比较小的区间，因为区间已经按照起点排序过，保留右端点值较小的区间能够尽量地

减小和往后扫描到的区间有重叠的概率。

/**
 * 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 int eraseOverlapIntervals(Interval[] intervals)
	{
		if(intervals.length<2)
			return 0;
		
		Arrays.sort(intervals,new Comparator<Interval>()
		{
			@Override
			public int compare(Interval o1, Interval o2)
			{
				// TODO Auto-generated method stub
				if(o1.start!=o2.start)
					return o1.start-o2.start;
				return o1.end-o2.end;
			}
		});
		
		Interval lastInterval=intervals[0];
		int cnt=0;
		
		for(int i=1;i<intervals.length;i++)
		{
			if(intervals[i].start<lastInterval.end)
			{
				cnt++;
				if(lastInterval.end<intervals[i].end)
					continue;
			}
			lastInterval=intervals[i];
		}
		
		return cnt;
	}
}