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 = 0; 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; } };
Next challenges: (M) Swap Nodes in Pairs (H) Copy List with Random Pointer (M) Insertion Sort List
时间: 2024-10-01 02:35:24