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.

 

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

求两个链表的第一个公共结点

思路一：遍历得到两个链表的长度，求出他们之差，用的长的链表先走若干步，接着在同时在两个链表上遍历，找到的第一个相同的结点就是他们的共同的结点

思路二：如果短链遍历完了还没有得到结果，则将指针指向长链头，长链头遍历完了之后指向短链头，这样来消除他们长度差异

C++（64ms）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int getListLen(ListNode* head){
        int len = 0 ;
        ListNode* p = head ;
        while(p != NULL){
            len++ ;
            p = p->next ;
        }
        return len ;
    }
    
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int len1 = getListLen(headA);
        int len2 = getListLen(headB);
        int lenDif = len1 - len2 ;
        ListNode* pLong = headA ;
        ListNode* pShort = headB ;
        if (len2 > len1){
            lenDif = len2 - len1 ;
            pLong = headB ;
            pShort = headA ;
        }
        while(lenDif--){
            pLong = pLong->next ;
        }
        while(pLong != pShort && pLong != NULL && pShort != NULL){
            pLong = pLong->next ;
            pShort = pShort->next ;
        }
        return pLong ;
    }
};

C++(37ms):

/**
 * 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* p1 = headA ;
        ListNode* p2 = headB ;
        if (p1 == NULL || p2 == NULL)
            return NULL ;
        while(p1 != NULL && p2 != NULL && p1 != p2){
            p1 = p1->next ;
            p2 = p2->next ;
            
            if (p1 == p2)
                return p1 ;
            if (p1 == NULL)
                p1 = headB ;
            if (p2 == NULL)
                p2 = headA ;
        }
        return p1 ;
    }
};