Python•intersection-of-two-linked-lists/synced-solution.py
class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
a = headA
b = headB
while a != b:
a = a.next if a else headB
b = b.next if b else headA
return a
C++•intersection-of-two-linked-lists/synced-solution.cpp
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* a = headA;
ListNode* b = headB;
while (a != b) {
a = a ? a->next : headB;
b = b ? b->next : headA;
}
return a;
}
};
TypeScript•intersection-of-two-linked-lists/synced-solution.ts
function getIntersectionNode(headA: ListNode | null, headB: ListNode | null): ListNode | null {
let a = headA;
let b = headB;
while (a !== b) {
a = a ? a.next : headB;
b = b ? b.next : headA;
}
return a;
}