[leetCode]160.相交链表
暴力法
每次从一个链表中取出一个结点,将该结点与另一个链表所有结点比对如果相等则返回当前结点。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode pA = headA;
ListNode pB = headB;
while(pA!=null){
while(pB!=null){
if(pA == pB) return pA;
pB = pB.next;
}
pA = pA.next;
pB = headB;
}
return null;
}
}
哈希表
将一个链表的元素加入哈希表,遍历另一个链表判断其结点是否存在于哈希表中。
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
HashSet<ListNode> set = new HashSet<>();
while(headA!=null && !set.contains(headA)){
set.add(headA);
headA = headA.next;
}
while(headB != null){
if(set.contains(headB)) return headB;
headB = headB.next;
}
return null;
}
}
双指针
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode pA = headA;
ListNode pB = headB;
if(pA == null || pB == null) return null;
while(pA !=null || pB != null){
if(pA == null) pA = headB;
else if(pB == null) pB = headA;
if(pA == pB) return pB;
pA = pA.next;
pB = pB.next;
}
return null;
}
}
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if(headA == null || headB == null) return null;
ListNode pA = headA;
ListNode pB = headB;
while(pA != pB){
pA = (pA == null ? headB : pA.next);
pB = (pB == null ? headA : pB.next);
}
return pB;
}
}