添加 面试题02.07.链表相交 python3版本

This commit is contained in:
evanlai
2021-05-30 14:21:41 +08:00
parent 013ba0575e
commit e98fb5a56f

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@ -151,7 +151,44 @@ public class Solution {
```
Python
```python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
lengthA,lengthB = 0,0
curA,curB = headA,headB
while(curA!=None): #求链表A的长度
curA = curA.next
lengthA +=1
while(curB!=None): #求链表B的长度
curB = curB.next
lengthB +=1
curA, curB = headA, headB
if lengthB>lengthA: #让curA为最长链表的头lenA为其长度
lengthA, lengthB = lengthB, lengthA
curA, curB = curB, curA
gap = lengthA - lengthB #求长度差
while(gap!=0):
curA = curA.next #让curA和curB在同一起点上
gap -= 1
while(curA!=None):
if curA == curB:
return curA
else:
curA = curA.next
curB = curB.next
return None
```
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