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.思路：贪心 算法，以每个元素的end为关键元素排序，end相等的情况下，按start升序排序。这里的实际类型和这个问题很像 非常重要的是对类进行排序
# Definition for an interval. # class Interval(object): # def __init__(self, s=0, e=0): # self.start = s # self.end = e class Solution(object): def eraseOverlapIntervals(self, intervals): if not intervals: return 0 intervals.sort(key=lambda x: x.start) res = 0 end = intervals[0].end for i in intervals[1:]: #print i.start,i.end if i.start < end: res += 1 end = min(end,i.end) else: end = i.end return res
