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.看了下别人的文章的思路，这是一道典型的贪心算法，类似于本科老师讲的安排活动的那道题。按照活动的结束时间进行排序，如果结束时间一样，那么开始时间早的放在前面。然后尽可能的保证安排的活动多（也就是剩余的少）。维护一个end变量，如果下一个活动的开始时间早于这个end，那么这个活动会被砍掉，否则的话更新end的值。
代码：
public int eraseOverlapIntervals(Interval[] intervals) { if(intervals == null || intervals.length == 0) return 0; Arrays.sort(intervals, new Comparator<Interval>() { @Override public int compare(Interval o1, Interval o2) { if(o1.end == o2.end){ return o1.start - o2.start; }else{ return o1.end - o2.end; } } }); int count = 0; int end = intervals[0].end; for(int i=1;i<intervals.length;i++){ if(intervals[i].start<end){ count++; } else{ end = intervals[i].end; } } return count; }思路参考： http://blog.csdn.net/liuchenjane/article/details/52972689
