[leetcode 160] Intersection of Two Linked Lists

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本文介绍了一种寻找两个单链表开始相交节点的方法。通过一次遍历确定交点是否存在，并找到具体的交点位置，算法力争达到O(n)的时间复杂度和O(1)的空间复杂度。

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.

/**
 * 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;
        }
        auto first = headA;
        auto second = headB;
        int ca = 1;
        int cb = 1;
        //判断是否有交叉
        while (first->next) {
            ca++;
            first = first->next;
        }
        while (second->next) {
            cb++;
            second = second->next;
        }
        if (first != second) {
            return NULL;
        }
        //求交叉点
        if (ca > cb) {
            first = headA;
            second = headB;
        } else {
            first = headB;
            second = headA;
        }
        for (int i = 0; i < abs(ca-cb); i++) {
            first = first->next;
        }
        while (first != second) {
            first = first->next;
            second = second->next;
        }
        return first;
        
    }
};