题目如下：
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: 1Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2], [1,2], [1,2] ]Output: 2Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,2], [2,3] ]Output: 0Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
解题思路：本题可以用贪心算法。 首先对 intervals 按start从小到大排序，如果start相同，则按end从小到大。接下来遍历intervals，如果intervals相邻的两个元素有重叠，删除掉end较大的那个，最后intervals中留下来的元素都是不重叠的。
代码如下：
# Definition for an interval.# class Interval(object):# def __init__(self, s=0, e=0):# self.start = s# self.end = eclass Solution(object): def eraseOverlapIntervals(self, intervals): """ :type intervals: List[Interval] :rtype: int """ def cmpf(i1,i2): if i1.start != i2.start: return i1.start - i2.start return i1.end - i2.end intervals.sort(cmp=cmpf) keep = None res = 0 while len(intervals) > 0: item = intervals.pop(0) if keep == None or keep.end <= item.start: keep = item elif keep.end <= item.end: res += 1 elif keep.end > item.end: keep = item res += 1 return res