统计删除几个区间使数组变为无重叠的区间 Non-overlapping Intervals

问题：

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:

1. You may assume the interval's end point is always bigger than its start point.
2. 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.

解决：

① 先排序，再删除。根据每一个区间的start来作升序排序，而后咱们开始要查找重叠区间，判断方法是看若是前一个区间的end大于后一个区间的start，那么必定是重复区间，此时咱们结果res自增1，咱们须要删除一个，那么此时咱们究竟该删哪个呢，为了保证咱们整体去掉的区间数最小，咱们去掉那个end值较大的区间。

/**
 * 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 { //5ms
    public int eraseOverlapIntervals(Interval[] intervals) {
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {//按照start升序排列
                return o1.start - o2.start;
            }
        });
        int count = 0;//记录要删除的个数
        int last = 0;
        int len = intervals.length;
        for (int i = 1;i < len;i ++){
            if (intervals[i].start < intervals[last].end){
                count ++;
                if (intervals[i].end < intervals[last].end) last = i;
            }else {
                last = i;
            }
        }
        return count;
    }
}