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.

Solution：Greedy Interative

思路：
这道题给了我们一堆区间，让我们求需要至少移除多少个区间才能使剩下的区间没有重叠，那么我们首先要给区间排序，根据每个区间的start来做升序排序，然后我们开始要查找重叠区间，判断方法是看如果前一个区间的end大于后一个区间的start，那么一定是重复区间
Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

/**
 * 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; }
 * }
 */
class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals.length == 0)  return 0;

        Arrays.sort(intervals, new myComparator());
        int end = intervals[0].end;
        int count = 1;        

        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start >= end) {
                end = intervals[i].end;
                count++;
            }
        }
        return intervals.length - count;
    }
    
    class myComparator implements Comparator<Interval> {
        public int compare(Interval a, Interval b) {
            return a.end - b.end;
        }
    }
}