[Leetcode] Intersection of Two Linked Lists - Go语言中文社区

[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.
题意:求得两个链表的交汇点;
思路:
1. 两个指针同时对两个链表分别遍历,当指针相同时说明存在交汇点
2. 当两个指针遍历到末尾且均不相同时说明不存在交汇点
3. 当指针A遍历至尾时,将其重定向至B链表头,继续遍历,相同地将B重定向至A,这样即便两个链表长度不一时,遍历速度快的那个指针总
能追上慢者

时间复杂度O(M+N),空间上使用了四个临时指针尴尬
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) {
		struct ListNode *pa = headA,*pb = headB;
        
        if(!pa || !pb)
        {
            return NULL;
        }
        
        struct ListNode *pa_last = NULL, *pb_last = NULL;
        
        while(pa != pb)
        {
            if(pa_last == NULL && pa->next == NULL)/* Last of PA */
            {
                pa_last = pa;
            }

            if(pb_last == NULL && pb->next == NULL)/* Last of PB */
            {
                pb_last = pb;
            }
            
            if(pa_last && pb_last && pa_last != pb_last)
            {
                return NULL;/* end of two list not equal,No intersection */
            }
            
            pa = pa->next;
            pb = pb->next;

            if(pa == NULL){
                pa = headB;
            }
            else if(pb == NULL){
                pb = headA;
            }
        }
        
        return pa;
}

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原文链接:https://blog.csdn.net/rudyn/article/details/53375425
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  • 发表于 2019-08-27 11:51:23
  • 阅读 ( 1462 )
  • 分类:算法

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