leetcode 435 无重叠区间

贪心想了半天，一直想不出完美的要解决条件，只知道要根据起点或终点排序。后来看了答案，原来是用总的区间数量减去没重复的区间数量，得到要去掉的区间数量。那就简单了，找出没重复的区间数量就好了，贪心，维护一段区间或者终点值就 ok 了。

class Solution(object):
    def eraseOverlapIntervals(self, intervals):
        """
        :type intervals: List[List[int]]
        :rtype: int
        """
        if len(intervals) <= 1:
            return 0
        intervals.sort(key=lambda x: x[1])
        end = intervals[0][1]
        count = 1
        for i in range(1, len(intervals)):
            if intervals[i][0] >= end:
                count += 1
                end = intervals[i][1]
        return len(intervals) - count