[leetcode] Intersection of Two Linked Lists

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本博客介绍了一个算法，用于找到两个单链表相交的起始节点。通过计算链表长度并调整指针位置，算法能够在O(n)时间复杂度下运行，仅使用O(1)内存。

From : https://leetcode.com/problems/intersection-of-two-linked-lists/

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 || !headB) return NULL;
        int len=0, lenb=0;
        ListNode* p = headA, *q = headB;
        while(p) {p=p->next; len++;}
        while(q) {q=q->next; lenb++;}
        if(len >= lenb) {p = headA; q = headB; len = len-lenb;}
        else {p = headB; q = headA; len = lenb-len;}
        while(len--) p = p->next;
        while(p != q) {
            p = p->next;
            q = q->next;
        }
        return p;
    }
};