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 visited = new HashSet();
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;
}



}

 


作者:一乐乐​