问题描述
RT.
解决思路
(1) 两链表都是单向链表：判断两链表的末尾节点是否相同；
(2) 两链表中一个有环，一个没环：不可能相交；
(3) 两链表都有环：slow-fast双指针方法。
程序
public class ListIntersection {// two single listpublic boolean isIntersectionOfTwoSingleList(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return false;}// whether the end of two list is sameListNode endOfList1 = getEndOfList(l1);ListNode endOfList2 = getEndOfList(l2);return endOfList1 == endOfList2;}private ListNode getEndOfList(ListNode head) {if (head == null) {return null;}ListNode node = head;while (node.next != null) {node = node.next;}return node;}// two list with cyclepublic boolean isIntersectionOfTwoListWithCycle(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return false;}ListNode slow = l1, fast = l2;while (fast.next != null || fast != null || slow != null) {slow = slow.next;fast = fast.next.next;if (slow == fast) {return true;}}return false;}
}
Follow up
求出两个链表相交的第一个节点（如果存在的话）。
(1) 两条单链表
a. 求出两条链表的长度及长度之差diff，然后设立两个指针指向两链表的头结点，其中指向长链表头结点的指针向前移动diff步；
b. 然后同时移动两指针，直到所指节点相同（地址相同）为止，否则返回null。
(2) 两条链表有环
首先slow-fast，直到相遇为止，其中任意一个指针指回其头结点，然后slow和fast指针同时移动，直到相遇，相遇的节点为第一个相交的节点。
（注意：可能有两个相交的节点）
程序
public class ListIntersection2 {// two single listpublic ListNode getFirstIntersectionNodeOfSingleList(ListNode l1,ListNode l2) {ListNode longNode = l1, shortNode = l2;int len1 = getLenOfList(l1);int len2 = getLenOfList(l2);if (len1 < len2) {longNode = l2;shortNode = l1;}int diff = Math.abs(len1 - len2);// long move diff stepswhile (diff > 0) {longNode = longNode.next;--diff;}while (longNode != null && shortNode != null) {if (longNode == shortNode) {return longNode;}longNode = longNode.next;shortNode = shortNode.next;}return null;}private int getLenOfList(ListNode head) {ListNode node = head;int len = 0;while (node != null) {++len;node = node.next;}return len;}// two list with cyclepublic ListNode getFirstIntersectionNodeOfCycleList(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return null;}ListNode slow = l1, fast = l2;while (fast.next != null || fast != null || slow != null) {if (fast == slow) {break;}slow = slow.next;fast = fast.next.next;}if (fast == null || slow == null) {return null;}slow = l1;while (slow != fast) {slow = slow.next;fast = fast.next;}return slow;}
}
)