LEETCODE: Intersection of Two Linked Lists

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本文介绍了一种在两个单链表中找到它们开始相交节点的方法。通过确定两个链表的长度并比较尾节点来判断是否存在交点，然后通过同步移动指针的方式找到交点位置。该方法适用于没有环的链表，并且能够在O(n)的时间复杂度内完成，仅使用O(1)的空间。

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.

/**
 * 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) {
        if(headA == NULL || headB == NULL)
            return NULL;
        
        // Find the length of list A and record the last node of list A.
        int lenA = 1;
        ListNode *lastA = headA;
        while(lastA != NULL && lastA->next != NULL) {
            lenA ++;
            lastA = lastA->next;
        }
        
        // Find the length of list B and record the last node of list B.
        int lenB = 1;
        ListNode *lastB = headB;
        while(lastB != NULL && lastB->next != NULL) {
            lenB ++;
            lastB = lastB->next;
        }
        
        // No intersaction.
        if(lastA != lastB)
            return NULL;
        
        // Skip the front nodes of the longer list, then move with the same speed.
        // The first time the two cursors meet is the intersaction node.
        lastA = headA;
        lastB = headB;
        if(lenA > lenB) {
            lenA -= lenB;
            
            for(int ii = 0; ii < lenA; ii ++) {
                lastA = lastA->next;
            }
        }
        else {
            lenB -=lenA;
            for(int ii = 0; ii < lenB; ii ++) {
                lastB = lastB->next;
            }
        }
        
        while(lastA != NULL && lastB != NULL && lastA != lastB) {
            lastA = lastA->next;
            lastB = lastB->next;
        }
        return lastA;
    }
};