160. Intersection of Two Linked Lists
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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.
找到相交的结点,很简单的题目,用双指针记录两个链表,如果相等则返回结点,如果没有则一直next下去,尽头没有next了就返回空。当其中一个指针走完的时候,就重定向到另外一个链表,这样子就可以让两个指针步数一致。在第二次迭代同时碰到交叉点结点。最后的时间复杂度是两个链表的长度之和,考虑最坏情况O(m+n)。例:A={1,2,3,4,5},B={6,7,8,3,4,5}。A的长度比B小,A先走完了,此时B走到4的位置,A重定向后在链表B的6的位置,此时B走到5,然后B重定向到A链表的位置1,最后两个指针距离交叉点3的步数一致。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) { 9 if(!headA || !headB) return NULL; 10 struct ListNode *pa=headA,*pb=headB; 11 while(1){ 12 if(pa==pb) 13 return pa; 14 else if(!pa->next && !pb->next) 15 return NULL; 16 else{ 17 if(pa->next) 18 pa=pa->next; 19 else 20 pa=headB; 21 if(pb->next) 22 pb=pb->next; 23 else 24 pb=headA; 25 } 26 } 27 }
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) { 9 if(!headA || !headB) return NULL; 10 struct ListNode *pa=headA,*pb=headB; 11 while(pa!=pb){ 12 pa=pa==NULL?headB:pa->next; 13 pb=pb==NULL?headA:pb->next; 14 } 15 return pa; 16 }
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