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例如：我有intervals = [[-5,-3],[-4,-1],[1,3],[4,8],[5,10],[10,12], [15,20]]（不必像那样排序）

我希望函数返回 me [[-5,-1],[1,3],[4,12],[15,20]]。因为[-5,-3],[-4,-1] and [4,8],[5,10],[10,12]有相互拦截的数字。即，我希望该函数返回所有“孤独”区间以及它们的数字相互截取的区间并集的最小值和最大值。

我有这段代码可以做类似的事情，但这还不是我想要的：

def maxDisjointIntervals(list_): 
      
    # Lambda function to sort the list   
    # elements by second element of pairs  
    list_.sort(key = lambda x: x[1]) 
      
    # First interval will always be  
    # included in set  
    print("[", list_[0][0], ", ", list_[0][1], "]") 
      
    # End point of first interval 
    r1 = list_[0][1] 
      
    for i in range(1, len(list_)): 
        l1 = list_[i][0] 
        r2 = list_[i][1] 
          
        # Check if given interval overlap with  
        # previously included interval, if not  
        # then include this interval and update  
        # the end point of last added interval 
        if l1 > r1: 
            print("[", l1, ", ", r2, "]") 
            r1 = r2

这段代码返回给我这个输出：[[-5, -3],[1, 3],[4, 8],[10, 12],[15, 20]]

重复我之前所说的：我希望它返回这个输出[[-5, -1],[1, 3],[4, 12],[15, 20]]

我希望我的解释不会太冗长，因为我不是以英语为母语的人。谢谢！

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2 回答 2

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所以我的理解是你想找到连接或重叠的区间，你可以通过使用迭代器来做到这一点。

def maxDisjointIntervals(intervals):  # dont use list_ as your variable name
    overlappedIntervals = []

    if len(intervals) == 0:
        return overlappedIntervals

    # sort the intervals using the starting time
    intervals = sorted(intervals, key=lambda interval: interval[0])

    # init the start hour and end hour
    startHour = intervals[0][0]
    endHour = intervals[0][1]

    for interval in intervals[1:]:
        # if there is an overlap
        if interval[0] <= endHour <= interval[1]:
            endHour = interval[1]
        # if there is not an overlap
        else:
            overlappedIntervals.append([startHour, endHour])
            startHour = interval[0]
            endHour = interval[1]
    overlappedIntervals.append([startHour, endHour])
    return overlappedIntervals


print(maxDisjointIntervals([[-5, -3], [-4, -1], [1, 3], [4, 8], [5, 10], [10, 12], [13, 20]]))

输出：[[-5, -1], [1, 3], [4, 12], [13, 20]]

于 2020-12-29T02:10:28.763 回答
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您可以使用 functools 中的 reduce 将间隔合并在一起：

intervals = [[-5,-3],[-4,-1],[1,3],[4,8],[5,10],[10,12], [15,20]]

from functools import reduce
disjoints = [*reduce(lambda a,b: a+[b] if not a or b[0]>a[-1][1] else a[:-1]+[[a[-1][0],b[1]]],intervals,[])]

print(disjoints) # [[-5, -1], [1, 3], [4, 12], [15, 20]]

或者在基本循环中做同样的事情：

disjoints = intervals[:1]
for s,e in intervals[1:]:
    if s>disjoints[-1][-1]: disjoints.append([s,e])
    else: disjoints[-1][-1] = e
    
print(disjoints) # [[-5, -1], [1, 3], [4, 12], [15, 20]]

注意：这假设包含范围。如果结束是独占使用>=而不是>.

于 2020-12-29T02:37:32.273 回答