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【leetcode】Intersection of Two Linked Lists(easy)

时间:2015-02-02 22:39:52      阅读:300      评论:0      收藏:0      [点我收藏+]

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.

思路:简单题 遍历长度 走差值 再同步走   答案里给了个转圈圈的思路 感觉没有自己的好 太乱了

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(headA == NULL || headB == NULL)
        {
            return NULL;
        }
        int lenA = 1;
        int lenB = 1;
        ListNode * pa = headA;
        ListNode * pb = headB;

        while(pa->next != NULL)
        {
            pa = pa->next;
            lenA++;
        }
        while(pb->next != NULL)
        {
            pb = pb->next;
            lenB++;
        }

        if(pa != pb) return NULL;

        pa = headA;
        pb = headB;
        while(lenA > lenB)
        {
            pa = pa->next; lenA--;
        }
        while(lenB > lenA)
        {
            pb = pb->next; lenB--;
        }
        while(pa != pb)
        {
            pa = pa->next;
            pb = pb->next;
        }

    }

 

【leetcode】Intersection of Two Linked Lists(easy)

原文:http://www.cnblogs.com/dplearning/p/4268754.html

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