【leetcode】435. Non-overlapping Intervals

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:

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.

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 = e

class 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

 

转载于:https://www.cnblogs.com/seyjs/p/10494587.html