package code; /* * 160. Intersection of Two Linked Lists * 题意：求两个链表的交叉点 * 难度：Easy * 分类：LinkedList * 思路：两种方法：1.找出两个链表的长度差x，长的先走x步； 2.走完一个链表，走另一个链表，两个cur都走了两个链表长度的和步 * Tips：两种方法的本质是一样的其实，都是找到了步数差，都遍历了两遍 */ public class lc160 { public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } public ListNode getIntersectionNode(ListNode headA, ListNode headB) { if(headA==null||headB==null) return null; ListNode curA = headA; ListNode curB = headB; while (curA != curB) { if(curA == null) curA = headB; if(curB == null) curB = headA; if(curA==curB) return curA; //别忘了判断一下是否相等 curA = curA.next; curB = curB.next; } return curA; } }