160. Intersection of Two Linked Lists

题目描述

Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:

begin to intersect at node c1.

方法

Approach 1: Brute Force

For each node ai in list A, traverse the entire list B and check if any node in list B coincides with ai.

Complexity Analysis

Time complexity : O(mn)

Space complexity : O(1)

Approach 2: Hash Table

Traverse list A and store the address / reference to each node in a hash set. Then check every node bi in list B: if bi appears in the hash set, then bi is the intersection node.

Complexity Analysis

Time complexity : O(m+n)

Space complexity : O(m)or O(n)

public class Solution { 
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        Set<ListNode> set = new HashSet<>();
        ListNode A = headA;
        ListNode B = headB;   //Time complexity : O(m+n). Space complexity : O(m) or O(n)
        while(A != null){
           if(!set.contains(A))
               set.add(A);
            A = A.next;
        }
        while( B != null){
            if(set.contains(B))
                return B;
            B = B.next;
        }
        return null;
    }
}

Approach 3: Two Pointers
思路分析：
查找两个链表的共同节点

要求：时间复杂度为 O(N)，空间复杂度为 O(1)

设 A 的长度为 a + c， B 的长度为 b + c，其中 c 为尾部公共部分长度，可知 a + c + b

= b + c + a。

当访问 A 链表的指针访问到链表尾部时，令它从链表 B 的头部开始访问链表 B；同样

地，当访问 B 链表的指针访问到链表尾部时，令它从链表 A 的头部开始访问链表 A。这样就能控制A和B两个链表的指针能同时访问到交点。
作者：花不休
来源：优快云
原文：https://blog.youkuaiyun.com/huabuxiu/article/details/81869013
版权声明：本文为博主原创文章，转载请附上博文链接！

public class Solution{
//Runtime: 1 ms, faster than 100.00% of Java 
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    //Approach 2: Two Pointers
    //boundary check
    if(headA == null || headB == null) return null;
    
    ListNode a = headA;
    ListNode b = headB;
    
    //if a & b have different len, then we will stop the loop after second iteration
    while( a != b){
    	//for the end of first iteration, 
    	//we just reset the pointer to the head of another linkedlist
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }
    
    return a;
}
}