Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
begin to intersect at node c1.
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3Output: Reference of the node with value = 8Input 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.
Example 2:
Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1Output: Reference of the node with value = 2Input 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.
Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2Output: nullInput 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.
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.
首先定义两个指针，分别遍历两个链表，得到链表长度。然后求长度差，得出两个链表的差值。开头定义的两个指针回到链表头部，让较长链表的指针先移动差值距离，然后两个指针一起向后移动，直到两个指针所指向的地址相同，即为所求交叉点。
/** * 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) { if(headA==NULL||headB==NULL) return NULL; ListNode *A =headA; ListNode *B =headB; int countA=0; int countB=0; int difference; while(A!=NULL) { A=A->next; countA++; } while(B!=NULL) { B=B->next; countB++; } A=headA; B=headB; if(countA>=countB) { difference=countA-countB; for(int i=0;A!=NULLi++) { A=A->next; } while(A!=NULL else { A=A->next; B=B->next; } } return NULL; } else { difference=countB-countA; for(int i=0;B!=NULLi++) { B=B->next; } while(B!=NULL else { A=A->next; B=B->next; } } return NULL; } }};