LeetCode 160之Intersection of Two Linked Lists 的Java题解

题目：

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2                   ↘                     c1 → c2 → c3                   ↗            B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

解答：

第一种解法：用哈希表存链表的节点判断是否有和之前相同的节点，时间空间都是o(m+n)

代码：

public static ListNode getIntersectionNode(ListNode headA, ListNode headB) {if(headA==null||headB==null)return null;ListNode cur1=headA;ListNode cur2=headB;Hashtable<ListNode, Integer> ht=new Hashtable<>();while(cur1!=null){ht.put(cur1, 1);cur1=cur1.next;}while(cur2!=null){if(ht.get(cur2)!=null){return cur2;}cur2=cur2.next;}return null;            }

第二种解法：利用链表头尾相接的方法，进行判断

代码：

public static ListNode getIntersectionNode2(ListNode headA, ListNode headB) {if(headA==null||headB==null)return null;ListNode curA=headA,curB=headB;while(curA!=curB){if(curA==null)curA=headB;elsecurA=curA.next;if(curB==null)curB=headA;elsecurB=curB.next;}return curA;}

第三种方法：

将两个链表末尾对齐，首先遍历两个链表得到长度，长度长的先走两个长度之差，后面同时走判断是否有相等的

代码：

public static ListNode getIntersectionNode3(ListNode headA, ListNode headB) {
if(headA==null||headB==null)
return null;
int lengthA=0,lengthB=0;
ListNode curA=headA,curB=headB;
while(curA!=null)
{
curA=curA.next;
lengthA++;
}
while(curB!=null)
{
curB=curB.next;
lengthB++;
}
curA=headA;
curB=headB;
if(lengthA>lengthB)
{
while(lengthA!=lengthB)
{
curA=curA.next;
lengthA--;
}
}
else if(lengthB>lengthA) {
while(lengthB!=lengthA)
{
curB=curB.next;
lengthB--;
}
}
while(curA!=null&&curB!=null)
{
if(curA==curB)
return curA;
else {
curA=curA.next;
curB=curB.next;
}
}
return null;

}