【leetcode】435. Non-overlapping Intervals【M】-优快云博客

本文链接：https://blog.youkuaiyun.com/sscssz/article/details/53213206

本文介绍了一个算法问题，即如何找出并移除一组区间中的最小数量的区间，使得剩下的区间不再相互重叠。通过示例说明了如何使用贪心算法解决该问题，并提供了Python实现代码。

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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