160. Intersection of Two Linked Lists

查找二个链表的第一个交叉节点:

首先查询二条链表的各自长度，其次长链表上先移动长度差的步数，然后同时遍历二个链表即可：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *A=headA;
        ListNode *B=headB;
        
        int len1=0,len2=0;
        while(headA!=NULL)
        {
            len1++;
            headA = headA->next;
        }
        while(headB!=NULL)
        {
            len2++;
            headB=headB->next;
        }
        
        if(len1>len2)
        {
            int len=len1-len2;
            while(len>0)
            {
                len--;
                A=A->next;
            }
        }
        if(len1<len2)
        {
            int len=len2-len1;
            while(len>0)
            {
                len--;
                B=B->next;
            }
        }
        
        
        while(A!=NULL&&B!=NULL)
        {
            if(A==B)
                return A;
            else
            {
                A=A->next;
                B=B->next;
            }
        }
        return NULL;
    } 
    
};