【leetcode】435. Non-overlapping Intervals【M】

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:

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.

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：貪心算法，以每個元素的end爲關鍵元素排序，end相等的情況下，按start升序排序。這裏的實際類型和這個問題很像

非常重要的是對類進行排序

# 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