答案 内容：
模板调用标记：
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其实思想可以按照从尾开始比较两个链表，如果相交，则从尾开始必然一致，只要从尾开始比较，直至不一致的地方即为交叉点，如图所示
# 使用a,b两个list来模拟链表，可以看出交叉点是 7这个节点
a = [1,2,3,7,9,1,5]
b = [4,5,7,9,1,5]
for i in range(1,min(len(a),len(b))):
if i==1 and (a[-1] != b[-1]):
print "No"
break
else:
if a[-i] != b[-i]:
print "交叉节点：",a[-i+1]
break
else:
pass
另外一种比较正规的方法，构造链表类
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def node(l1, l2):
length1, lenth2 = 0, 0
# 求两个链表长度
while l1.next:
l1 = l1.next
length1 += 1
while l2.next:
l2 = l2.next
length2 += 1
# 长的链表先走
if length1 > lenth2:
for _ in range(length1 - length2):
l1 = l1.next
else:
for _ in range(length2 - length1):
l2 = l2.next
while l1 and l2:
if l1.next == l2.next:
return l1.next
else:
l1 = l1.next
l2 = l2.next
修改了一下:
#coding:utf-8
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def node(l1, l2):
length1, length2 = 0, 0
# 求两个链表长度
while l1.next:
l1 = l1.next#尾节点
length1 += 1
while l2.next:
l2 = l2.next#尾节点
length2 += 1
#如果相交
if l1.next == l2.next:
# 长的链表先走
if length1 > length2:
for _ in range(length1 - length2):
l1 = l1.next
return l1#返回交点
else:
for _ in range(length2 - length1):
l2 = l2.next
return l2#返回交点
# 如果不相交
else:
return
思路: http://humaoli.blog.163.com/blog/static/13346651820141125102125995/ |