linked list 8

1. Topic

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.

2. Answer: The meaning of the question is to find the common starting node of the two linked lists

3. C++ code

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *pA = headA,*pB = headB;
        int lenA = 0,lenB = 0;
        while(pA){
            lenA++;
            pA = pA->next;
        }
        while(pB){
            lenB++;
            pB = pB->next;
        }
        pA = headA,pB = headB;
        
        if(lenA > lenB){
            for(int i =0;i<lenA-lenB;i++)
                pA = pA->next;
        }else{
            for(int i=0;i<lenB-lenA;i++)
                pB = pB->next;
        }
        
        while(pA){
            if(pA == pB)
                return pA;
            pA = pA->next;
            pB = pB->next;
        }
        return NULL;
       
    }
};

python code

class Solution(object):
    def getIntersectionNode (self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        curA,curB = headA,headB
        lenA, lenB = 0.0
        while curA is not None:
            lenA += 1
            curA = curA.next
        while curB is not None:
            lenB += 1
            curB = curB.next
        curA,curB = headA,headB
        if lenA > lenB:
            for i in range(lenA-lenB):
                curA = curA.next
        elif lenB> lenA:
            for i in range(lenB-lenA):
                curB = curB.next
        while curB != curA:
            curB = curB.next
            curA = curA.next
        return curA