Total Accepted: 53721 Total Submissions: 180705 Difficulty: Easy
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
- 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.
- Your code should preferably run in O(n) time and use only O(1) memory.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int getListLength(ListNode* head){
int len = ;
while(head){
len++;
head=head->next;
}
return len;
}
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
int lenA = getListLength(headA);
int lenB = getListLength(headB);
ListNode* common = NULL,*startA=headA,*startB=headB;
if(lenA<lenB){
int diff = lenB-lenA;
while(diff--) startB=startB->next;
}else{
int diff = lenA-lenB;
while(diff--) startA=startA->next;
}
while(startA){
if(startA == startB){
common = startA;
break;
}
startA=startA->next;
startB=startB->next;
}
return common;
}
};