[leetcode]#160 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.

（1）用O(1)空间和O(n)时间，找出两条链表公共部分的起点

解题思路：1、计算两者长度，并判断两个链表尾端是否相同，不同返回null

2、让链表长度较长的走到 （tail的数量 - Btail的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点

[code]class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode * tail = headA;
        ListNode * Btail = headB;
        int countA = 1, countB = 1, step;
        if (tail == NULL || headB == NULL)
            return NULL;
        while (tail->next != NULL) {
            tail = tail->next;
            countA++;
        }
        while (Btail->next != NULL) {
            Btail = Btail->next;
            countB++;
        }
        if (Btail != tail)
            return NULL;
        step = countA - countB;
        tail = headA;
        Btail = headB;
        if (step < 0) {
            step = -step;
            tail = headB;
            Btail = headA;
        }
        for (int i = 0;i < step; i++) {
            tail = tail->next;
        }
        while (tail != NULL && Btail != NULL && tail !=Btail) {
            tail = tail->next;
            Btail = Btail->next;
        }
        return tail;
    }
};