[LeetCode]160. Intersection of Two Linked Lists

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原题链接:https://leetcode.com/problems/contains-duplicate-iii/

这道题目是easy题,所以会有可能出现在电面里面。思路也很简单,设置2个指针分别指向2个链表,如果链表长度不相等,假设2个链表相差k个节点,则我们让长度长的那个链表对应的指针先走前进k个节点,然后两个指针一起走,直走到2个指针指向相同的节点为止。时间复杂度是O(n)

代码如下:

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if(headA == null || headB == null)            return null;        int lenA = 0;        ListNode pA = headA;        int lenB = 0;        ListNode pB = headB;        while(pA!=null)        {            lenA++;            pA = pA.next;        }        while(pB!=null)        {            lenB++;            pB = pB.next;        }        int diff = Math.abs(lenA-lenB);        pA = headA;        pB = headB;        if(lenA>lenB)        {            while(diff>0 && pA!=null)            {                pA = pA.next;                diff--;            }        }        else if(lenA<lenB)        {            while(diff>0 && pB!=null)            {                pB = pB.next;                diff--;            }        }        while(pA!=null && pB!=null)        {            if(pA == pB)                return pA;            pA = pA.next;            pB = pB.next;        }        return null;    } 


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