链表8

绝地灬酷狼 2022-05-25 01:15 83阅读 0赞

1、题目

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、解答：题目的意思是 找到两个链表的共同起始结点

3、C++代码

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代码

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