[LeetCode] Intersection of Two Linked Lists
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.
Credits:
Special thanks to @stellari for adding this problem and creating all test cases.
Linked List
分析:
如果两个链表相交,那么后面的链表必然相同,找到lenA,lenB,然后跳过lenA和lenB的差,
然后比较两个指针是否相同,若相同则有交叉,同时找到交叉节点,否则就是没有交叉。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { if(headA == NULL || headB == NULL) return NULL; int lenA = 0; int lenB = 0; ListNode* pA = headA; ListNode* pB = headB; while(pA) { lenA ++; pA = pA->next; } while(pB) { lenB ++; pB = pB->next; } if(lenA > lenB) { for(int i = 0; i < (lenA-lenB); i++) headA = headA->next; } else if(lenA < lenB) { for(int i = 0; i < (lenB-lenA); i++) headB = headB->next; } while(headA != headB) { headA = headA->next; headB = headB->next; } if(headA == headB) return headA; else return NULL; } };
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