Leetcode 435 python 解题报告

AC代码：

class Solution(object):
    def eraseOverlapIntervals(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        if len(intervals) == 0:
            return 0
        intervals = sorted(intervals, key=lambda Interval: Interval.start)
        prev = 0
        count = 0
        for i in range(1,len(intervals)):
            if intervals[prev].end > intervals[i].start:
                if intervals[prev].end > intervals[i].end:
                    prev = i
                count += 1
            else:
                prev = i
        return count

解题思路：自定义排序函数，按interval的start排序，然后分情况讨论，如果后数与前数是包含关系，则贪心法直接舍弃前数，如果是交的关系，则贪心法舍弃后数。