LeetCode——160 Intersection of Two Linked Lists

题目

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3

Output: Reference of the node with value = 8

Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1

Output: Reference of the node with value = 2

Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2

Output: null

Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.

Explanation: The two lists do not intersect, so return null.

代码

class Solution {

public:

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

        int lenA = getLen(headA), lenB = getLen(headB);

        if(lenA>lenB){

            for(int i=0; i<lenA-lenB; i++){

                headA = headA->next;

            }

        }

        else{

            for(int i=0; i<lenB-lenA; i++){

                headB = headB->next;

            }

        }

        while(headA&&headB&&headA!=headB){

            headA = headA->next;

            headB = headB->next;

        }

        return (headA==headB) ? headA : NULL;

    }

private:

    int getLen(ListNode *list){

        int cnt = 0;

        ListNode *tmp = list;

        while(tmp){

            tmp = tmp->next;

            ++cnt;

        }

        return cnt;

    }

};

思路

先计算出两个链表的长度，然后进行比较，将较长的链表缩短（即将头节点指针向后移），使得两个链表长度一致，然后让指针同时同步长迭代，当发现地址相同时则知道当前节点开始共用。

巴特西

LeetCode——160 Intersection of Two Linked Lists

题目

代码

思路

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