380. Intersection of Two Linked Lists【medium】
Write a program to find the node at which the intersection of two singly linked lists begins.
Notice
- If the two linked lists have no intersection at all, return
null
. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
The following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Your code should preferably run in O(n) time and use only O(1) memory.
解法一:
class Solution {
public:
/**
* @param headA: the first list
* @param headB: the second list
* @return: a ListNode
*/
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
// write your code here
if(headA == NULL || headB == NULL)
return NULL;
ListNode* iter1 = headA;
ListNode* iter2 = headB;
int len1 = ;
while(iter1->next != NULL)
{
iter1 = iter1->next;
len1 ++;
}
int len2 = ;
while(iter2->next != NULL)
{
iter2 = iter2->next;
len2 ++;
}
if(iter1 != iter2)
return NULL;
if(len1 > len2)
{
for(int i = ; i < len1-len2; i ++)
headA = headA->next;
}
else if(len2 > len1)
{
for(int i = ; i < len2-len1; i ++)
headB = headB->next;
}
while(headA != headB)
{
headA = headA->next;
headB = headB->next;
}
return headA;
}
};
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