435. Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

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.

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.

分析：

将区间按照尾部大小sort，然后挨个去掉overlap的区间，这样留下来的就是占用最小位置的区间。貌似做过这道题，没想到是贪心算法。

Solution：

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])
 
 ans = 0
 prev = intervals[0][1]
 
 for i in range(1, len(intervals)):
   if intervals[i][0]< prev:
     ans += 1
   else:
     prev = intervals[i][1]
 return ans