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 = 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.
Example 3:
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.
方法一 先分别计算两个链表的长度。先遍历较长的链表，走到两个链表等长的时候再同时走两个链表，并判断两个链表遍历到的当前节点是否相同。
public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode curA = headA; ListNode curB = headB; int lenA = 0; int lenB = 0; while (curA != null) { lenA++; curA = curA.next; } while (curB != null) { lenB++; curB = curB.next; } curA = headA; curB = headB; if (lenA > lenB) { int i = lenA - lenB; while (i != 0) { curA = curA.next; i--; } while (curA != null && curB != null) { if (curA == curB) { return curA; } curA = curA.next; curB = curB.next; } } else { int i = lenB - lenA; while (i != 0) { curB = curB.next; i--; } while (curA != null && curB != null) { if (curA == curB) { return curA; } curA = curA.next; curB = curB.next; } } return null; } }方法二： 利用set，存储链表一的节点。然后遍历链表B，如果找到set中有这个节点，则是交点，返回。
public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode curA = headA; ListNode curB = headB; Set<ListNode> set = new HashSet<>(); while (curA != null) { set.add(curA); curA = curA.next; } while (curB != null) { if (set.contains(curB)) { return curB; } curB = curB.next; } return null; } }