(**leetcode_LinkedList) Intersection of Two Linked Lists

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本文介绍了一种使用双指针计算长度的方法来找到两个单链表开始相交的节点。该方法首先分别计算两个链表的长度，然后让较长的链表先前进长度之差步数，之后两个链表同时前进直到它们相遇或者都到达末尾。

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Intersection of Two Linked Lists

Total Accepted: 22543 Total Submissions: 81453 My Submissions

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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

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双指针计算长度法

/**
 * 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) {
        int lenA = 0, lenB = 0;
        ListNode *tmpA = headA, *tmpB = headB;
        while(NULL!=tmpA){
            lenA++;
            tmpA = tmpA->next;
        }
        while(NULL!=tmpB){
            lenB++;
            tmpB = tmpB->next;
        }
        if(lenB>lenA)
            for(int i=0;i<(lenB-lenA); i++)
                headB=headB->next;
        else
            for(int i=0;i<(lenA-lenB); i++)
                headA=headA->next;
        while(headA!=headB){
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
        
    }
};