[LeetCode] 160. Intersection of Two Linked Lists 解题思路

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.

问题：判断两个列表是否有相交的元素，若有，找出相交节点。

这题也是一道基础题，看了自己的列表知识还需要巩固下才好。

分别求出两个列表的长度 len1, len2 ，以及他们的长度差异 diff

跳过长度差异部分，对于剩余的相同长度部分，依次检查两个链表的对应节点，若又存在相交节点，则必有两个对应节点相等。

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        
        int len1 = 0;
        ListNode* p1 = headA;
        while(p1 != NULL){
            p1 = p1->next;
            len1++;
        }
        
        int len2 = 0;
        ListNode* p2 = headB;
        while(p2 != NULL){
            p2 = p2->next;
            len2++;
        }
        
        p1 = headA;
        p2 = headB;
        if (len1 > len2){
            int diff = len1 - len2;
            while(diff > 0){
                p1 = p1->next;
                diff--;
            }
        }
        
        if (len2 > len1){
            int diff = len2 - len1;
            while(diff > 0){
                p2 = p2->next;
                diff--;
            }
        }
        
        while(p1 != NULL ){
            if ( p1 == p2){
                return p1;
            }
            p1 = p1->next;
            p2 = p2->next;
        }
        
        return NULL;
    }

参考资料：

LeetCode: Intersection of Two Linked Lists 解题报告, Yu's garden