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 }

     

    看到别的大神更简洁的写法,同样的原理:
     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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