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。