https://www.codewars.com/kata/sum-of-intervals/train/python
Write a function called sumIntervals/sum_intervals() that accepts an array of intervals, and returns the sum of all the interval lengths. Overlapping intervals should only be counted once.
Intervals
Intervals are represented by a pair of integers in the form of an array. The first value of the interval will always be less than the second value. Interval example: [1, 5] is an interval from 1 to 5. The length of this interval is 4.
Overlapping Intervals
List containing overlapping intervals:
[
[1,4],
[7, 10],
[3, 5]
]The sum of the lengths of these intervals is 7. Since [1, 4] and [3, 5] overlap, we can treat the interval as [1, 5], which has a length of 4.
Examples:
sumIntervals( [
[1,2],
[6, 10],
[11, 15]
] ); // => 9
sumIntervals( [
[1,4],
[7, 10],
[3, 5]
] ); // => 7
sumIntervals( [
[1,5],
[10, 20],
[1, 6],
[16, 19],
[5, 11]
] ); // => 19普通的思路：
先用双指针合并区间，然后再遍历合并好的区间统计总长度。
时间复杂度：O（NlogN）
空间复杂度：O（1）
def sum_of_intervals(intervals):
if not intervals:
return 0
def merge_intervals(intervals):
intervals.sort()
res = []
start, end = intervals[0][0], intervals[0][1]
for i, interval in enumerate(intervals[1:]):
s, e = interval[0], interval[1]
if s > end:
res.append([start, end])
start, end = s, e
else:
end = max(e, end)
res.append([start, end])
return res
intervals = merge_intervals(intervals)
res = 0
for s, e in intervals:
res += e - s
return res奇淫巧技的思路：
把所有区间里的数丢到一个集合里，然后统计集合的长度即可。
时间复杂度：O（N），N是intervals长度
空间复杂度：O（N*M），M是最长的那个interval的长度
def sum_of_intervals(intervals):
l = []
for s, e in intervals:
l += range(s, e)
return len(set(l))或者Codewars经典的一行解。
def sum_of_intervals(intervals):
return len(set(i for s, e in intervals for i in range(s, e)))