160. Intersection of Two Linked Lists

问题描述：

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.

 

解题思路：

如果两个linkedlist有相交的地方，纳闷他们的后半部分是相同的。

这说明我们可以开始从登长的地方开始搜索。

将长的链表的指针移动到与短链表长度相同的起始处，开始比较是否相同。

 

代码：

/**
 * 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 lenA = getLen(headA);
        int lenB = getLen(headB);
        ListNode* curA = headA;
        ListNode* curB = headB;
        if(lenA > lenB){
            int diff = lenA - lenB;
            while(diff > 0){
                curA = curA->next;
                diff--;
            }
        }else if(lenA < lenB){
            int diff = lenB - lenA;
            while(diff > 0){
                curB = curB->next;
                diff--;
            }
        }
        while(curA && curB){
            if(curA == curB)
                return curA;
            curA = curA->next;
            curB = curB->next;
        }
        return NULL;
    }
private:
    int getLen(ListNode *head){
        int ret = 0;
        ListNode *cur = head;
        while(cur){
            ret++;
            cur = cur->next;
        }
        return ret;
    }
};