文章作者：Tyan
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1. Description
Non-overlapping Intervals
2. Solution
class Solution:
def eraseOverlapIntervals(self, intervals):
count = 0
if len(intervals) < 2:
return count
intervals.sort(key=lambda interval: interval[1])
overlap = intervals[0]
for index in range(1, len(intervals)):
if overlap[1] > intervals[index][0]:
count += 1
else:
overlap = intervals[index]
return count
