我用Python实现了它。主要思路类似于归并排序。在
lists中有k个数组。在函数mainMerageK中,只需将列表(k)分成左侧(k/2)和右侧(k/2)。因此,划分的总计数为log(k)。关于函数merge,很容易知道运行时间为O(n)。最后,我们得到O(nlog k)。
顺便说一下,它也可以在最小堆中实现,这里有一个链接:
使用优先队列合并K个已排序列表。
def mainMergeK(*lists):
k = len(lists)
if k > 1:
mid = int(k / 2)
B = mainMergeK(*lists[0: mid])
C = mainMergeK(*lists[mid:])
A = merge(B, C)
print B, ' + ', C, ' = ', A
return A
return lists[0]
def merge(B, C):
A = []
p = len(B)
q = len(C)
i = 0
j = 0
while i < p and j < q:
if B[i] <= C[j]:
A.append(B[i])
i += 1
else:
A.append(C[j])
j += 1
if i == p:
for c in C[j:]:
A.append(c)
else:
for b in B[i:]:
A.append(b)
return A
if __name__ == '__main__':
x = mainMergeK([1, 3, 5], [2, 4, 6], [7, 8, 10], [9])
print x
输出结果如下:
[1, 3, 5] + [2, 4, 6] = [1, 2, 3, 4, 5, 6]
[7, 8, 10] + [9] = [7, 8, 9, 10]
[1, 2, 3, 4, 5, 6] + [7, 8, 9, 10] = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]