LeetCode 160. Intersection of Two Linked Lists(C++版)

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.

思路分析：

可以将A,B两个链表看做两部分，交叉前与交叉后。

交叉后的长度是一样的，因此交叉前的长度差即为总长度差diff。

只要去除这些长度差，距离交叉点就等距了。

然后先让较长的链表走diff步。然后两个链表开始同时走，若指针相等，则为交叉点的开始。

/**
 * 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) {
        ListNode *cpA = headA;
        ListNode *cpB = headB;
        int cntA = 0, cntB = 0;
        while(cpA){
            cntA ++;
            cpA = cpA -> next;
        }
        while(cpB){
            cntB ++;
            cpB = cpB -> next;
        }
        cpA = headA;
        cpB = headB;
        bool a_is_longer = true;
        if(cntA < cntB)
            a_is_longer = false;
        int diff = abs(cntA-cntB);
        
        if(a_is_longer){
            for(; diff > 0; diff --){
                cpA = cpA -> next;
            }
        }
        else{
            for(; diff > 0; diff --){
                cpB = cpB -> next;
            }
        }
        
        while(cpA){
            if(cpA == cpB) return cpA;
            cpA = cpA -> next;
            cpB = cpB -> next;
        }
        return cpA;
    }
};