160_相交链表
160_相交链表
package 链表; import java.util.HashSet; import java.util.Set; /** * https://leetcode-cn.com/problems/intersection-of-two-linked-lists/ * @author Huangyujun * */ public class _160_相交链表 { //方法一:Set集合(装入一条链表,然后以它为标准,依次拿另外一条链表的每个结点与它对比) public ListNode getIntersectionNode(ListNode headA, ListNode headB) { Set<ListNode> visited = new HashSet<ListNode>(); ListNode temp = headA; while (temp != null) { visited.add(temp); temp = temp.next; } temp = headB; while (temp != null) { if (visited.contains(temp)) { return temp; } temp = temp.next; } return null; } //方法二:处理一下长度,使得两条链表的长度相同后进行同步运动: public ListNode getIntersectionNode2(ListNode headA, ListNode headB) { ListNode curA = headA; ListNode curB = headB; int lenA = 0, lenB = 0; while (curA != null) { // 求链表A的长度 lenA++; curA = curA.next; } while (curB != null) { // 求链表B的长度 lenB++; curB = curB.next; } curA = headA; curB = headB; // 让curA为最长链表的头,lenA为其长度 if (lenB > lenA) { //1. swap (lenA, lenB); int tmpLen = lenA; lenA = lenB; lenB = tmpLen; //2. swap (curA, curB); ListNode tmpNode = curA; curA = curB; curB = tmpNode; } // 求长度差 int gap = lenA - lenB; // 让curA和curB在同一起点上(末尾位置对齐) while (gap-- > 0) { curA = curA.next; } // 遍历curA 和 curB,遇到相同则直接返回 while (curA != null) { if (curA == curB) { return curA; } curA = curA.next; curB = curB.next; } return null; } }
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