检查时间段是否重叠

Question

我有一个时间记录的列表，时间格式是HHMM（小时和分钟）。每个记录都有一个开始时间和一个结束时间。我在用Python编程时遇到了麻烦，想要判断这个列表中的时间是否有重叠。

举个例子

Entry 1: 1030, 1245;
Entry 2: 1115, 1300
== True

Entry 1: 0900, 1030;
Entry 2: 1215, 1400
== False

Answer 1

为了进一步解释@Roy的回答，特别是在某些情况下，当多个事件在同一个时间段内发生时，你需要区分它们：

intervals = [[100,200, "math"],[100,200, "calc"], [150,250, "eng"],[300,400, "design"],[250,500, "lit"],[10,900, "english"],[1000,12300, "prog"],[-151,32131, "hist"]]

overlapping = [ [x,y] for x in intervals for y in intervals if x is not y and x[1]>y[0] and x[0]<y[0] or x[0]==y[0] and x[1]==y[1] and x is not y]
for x in overlapping:
    print '{0} overlaps with {1}'.format(x[0],x[1])


# Prints
#[100, 200, 'math'] overlaps with [100, 200, 'calc']
#[100, 200, 'math'] overlaps with [150, 250, 'eng']
#[100, 200, 'calc'] overlaps with [100, 200, 'math']
#[100, 200, 'calc'] overlaps with [150, 250, 'eng']
#[250, 500, 'lit'] overlaps with [300, 400, 'design']
#[10, 900, 'english'] overlaps with [100, 200, 'math']
#[10, 900, 'english'] overlaps with [100, 200, 'calc']
#[10, 900, 'english'] overlaps with [150, 250, 'eng']
#[10, 900, 'english'] overlaps with [300, 400, 'design']
#[10, 900, 'english'] overlaps with [250, 500, 'lit']
#[-151, 32131, 'hist'] overlaps with [100, 200, 'math']
#[-151, 32131, 'hist'] overlaps with [100, 200, 'calc']
#[-151, 32131, 'hist'] overlaps with [150, 250, 'eng']
#[-151, 32131, 'hist'] overlaps with [300, 400, 'design']
#[-151, 32131, 'hist'] overlaps with [250, 500, 'lit']
#[-151, 32131, 'hist'] overlaps with [10, 900, 'english']
#[-151, 32131, 'hist'] overlaps with [1000, 12300, 'prog']

Answer 2

为了以后参考，@Roy 提出的解决方案不适用于那些开始时间或结束时间相同的区间。下面的解决方案可以解决这个问题：

intervals = [[100, 200], [150, 250], [300, 400], [250, 500], [100, 150], [175, 250]]
intervals.sort()
l = len(intervals)
overlaps = []
for i in xrange(l):
  for j in xrange(i+1, l):
    x = intervals[i]
    y = intervals[j]
    if x[0] == y[0]:
      overlaps.append([x, y])
    elif x[1] == y[1]:
      overlaps.append([x, y])
    elif (x[1]>y[0] and x[0]<y[0]):
      overlaps.append([x, y])

另外，可以使用区间树来处理这类问题。

Answer 3

首先，我们按照开始时间对列表进行排序。

然后，我们遍历这个列表，检查下一个开始时间是否小于上一个结束时间。

这样做是为了检查第x+1个时间段是否与第x个时间段重叠（而不是检查第x+2个时间段是否与第x个时间段重叠，等等）。

intervals = [[100,200],[150,250],[300,400]]
intervalsSorted = sorted(intervals, key=lambda x: x[0]) # sort by start time
for x in range(1,len(intervalsSorted)):
    if intervalsSorted[x-1][1] > intervalsSorted[x][0]:
        print "{0} overlaps with {1}".format( intervals[x-1], intervals[x] )

# result: [100, 200] overlaps with [150, 250]

接下来，这个方法应该能给你整个列表中所有重叠的时间段。

intervals = [[100,200],[150,250],[300,400],[250,500]]

overlapping = [ [x,y] for x in intervals for y in intervals if x is not y and x[1]>y[0] and x[0]<y[0] ]
for x in overlapping:
    print '{0} overlaps with {1}'.format(x[0],x[1])

# results:
# [100, 200] overlaps with [150, 250]
# [250, 500] overlaps with [300, 400]

需要注意的是，这个查找的复杂度是O(n*n)。（如果我说错了，请大家纠正我！）

这个方法可能比第一个方法要慢（我没有测试过，但我猜是这样），因为它会对每一个索引都遍历整个列表。这个过程应该和arbarnert的嵌套循环示例类似。不过，这个方法确实能给你所有重叠的值，而不是像我之前展示的第一个方法那样，只检查相邻的时间段（按照开始时间排序的）。

扩展测试结果是：

intervals = [[100,200],[150,250],[300,400],[250,500],[10,900],[1000,12300],[-151,32131],["a","c"],["b","d"],["foo","kung"]]

overlapping = [ [x,y] for x in intervals for y in intervals if x is not y and x[1]>y[0] and x[0]<y[0] ]
for x in overlapping:
    print '{0} overlaps with {1}'.format(x[0],x[1])

# results:
# [100, 200] overlaps with [150, 250]
# [250, 500] overlaps with [300, 400]
# [10, 900] overlaps with [100, 200]
# [10, 900] overlaps with [150, 250]
# [10, 900] overlaps with [300, 400]
# [10, 900] overlaps with [250, 500]
# [-151, 32131] overlaps with [100, 200]
# [-151, 32131] overlaps with [150, 250]
# [-151, 32131] overlaps with [300, 400]
# [-151, 32131] overlaps with [250, 500]
# [-151, 32131] overlaps with [10, 900]
# [-151, 32131] overlaps with [1000, 12300]
# ['a', 'c'] overlaps with ['b', 'd']