合併兩個排序的鍊錶

題目描述：

輸入兩個單調遞增的鍊錶，輸出兩個鍊錶合成後的鍊錶，當然我們需要合成後的鍊錶滿足單調不減規則。

答案：非遞迴

# -*- coding:utf-8 -*-

# class listnode:
# def __init__(self, x):
# self.val = x
# self.next = none
class solution:
# def merge(self,phead1, phead2):
# write code here
if phead1==none and phead2==none:
return none
if phead1==none:
return phead2
if phead2==none:
return phead1
phead3 = listnode(0)
p1 = phead1
p2 = phead2
if phead1.val < phead2.val:
phead3 = phead1
p1 = phead1.next
else:
phead3 = phead2
p2 = phead2.next
p3 = phead3#每次指向要新合併的節點
while p1 and p2:
if p1.val < p2.val:
p3.next = p1
p1 = p1.next
else:
p3.next = p2
p2 = p2.next
p3 = p3.next
if p1:
p3.next = p1
else:
p3.next = p2
return phead3

遞迴

def merge(phead1, phead2):

# write code here
if phead1 == none:
return phead2
elif phead2 == none:
return phead1
if phead1.val <= phead2.val:
phead1.next = merge(phead1.next, phead2)
return phead1
else:
phead2.next = merge(phead1, phead2.next)
return phead2

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