合并两个排序数组
def mergeList(A, B):
s1 = len(A)
s2 = len(B)
i,j = 0,0
res = []
while i < s1 and j < s2:
if A[i] <= B[j]:
res.append(A[i])
i += 1
else:
res.append(B[j])
j += 1
res = res + A[i+1:] + B[j+1:]
return res
A = [1,2,5,7,9]
B = [2,4,6,8,10,11,34,55]
res = mergeList(A, B)
print(res)
合并多个有序列表
def mergeMultiList(lists):
import heapq
from collections import deque
lists = list(map(lambda x: deque(x), lists))
pq = []
for ind, val in enumerate(lists):
pq.append((val.popleft(), ind))
heapq.heapify(pq)
res = []
while pq:
value, index = heapq.heappop(pq)
print(value, index)
res.append(value)
if lists[index]:
heapq.heappush(pq, (lists[index].popleft(), index))
return res
lists = [[1,2,5,7,9],[2,4,6,8,10,11,34,55],[1,3,5,8,10,15]]
res = mergeMultiList(lists)
print(res)
寻找两个有序列表中的中位数
class Solution:
"""
@param A: An integer array.
@param B: An integer array.
@return: a double whose format is *.5 or *.0
"""
def findMedianSortedArrays(self, A, B):
n = len(A) + len(B)
if n % 2 == 1:
return self.findKth(A, B, n / 2 + 1)
else:
smaller = self.findKth(A, B, n / 2)
bigger = self.findKth(A, B, n / 2 + 1)
return (smaller + bigger) / 2.0
def findKth(self, A, B, k):
if len(A) == 0:
return B[int(k - 1)]
if len(B) == 0:
return A[int(k - 1)]
if k == 1:
return min(A[0], B[0])
a = A[int(k / 2) - 1] if len(A) >= k / 2 else None
b = B[int(k / 2) - 1] if len(B) >= k / 2 else None
if b is None or (a is not None and a < b):
return self.findKth(A[int(k / 2):], B, int(k - k // 2))
return self.findKth(A, B[int(k / 2):], int(k - k // 2))
s = Solution()
print(s.findMedianSortedArrays([1, 2, 3, 4, 5, 6], [2, 3, 4, 5]))
