[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)空間和O(n)時間，找出兩條鏈表公共部分的起點

解題思路：1、計算兩者長度，並判斷兩個鏈表尾端是否相同，不同返回null
2、讓鏈表長度較長的走到 （tail的數量 - Btail的數量）步，然後和短鏈表同步往下走，遇到的第一個相同的節點就是最早的公共節點

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;
    }
};