（Java）leetcode-160 Intersection of Two Linked Lists

迷南。 2022-04-24 01:34 90阅读 0赞

# 题目 #

【寻找两条单链表的交叉点】  
Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3o3MTQ0MDU0ODk_size_16_color_FFFFFF_t_70]

begin to intersect at node c1.

Example 1:  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3o3MTQ0MDU0ODk_size_16_color_FFFFFF_t_70 1]

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
    Output: Reference of the node with value = 8

Input Explanation: The intersected node’s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as \[4,1,8,4,5\]. From the head of B, it reads as \[5,0,1,8,4,5\]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3o3MTQ0MDU0ODk_size_16_color_FFFFFF_t_70 2]

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
    Output: Reference of the node with value = 2

Input Explanation: The intersected node’s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as \[0,9,1,2,4\]. From the head of B, it reads as \[3,2,4\]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3o3MTQ0MDU0ODk_size_16_color_FFFFFF_t_70 3]

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
    Output: null

Input Explanation: From the head of A, it reads as \[2,6,4\]. From the head of B, it reads as \[1,5\]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.  
Explanation: The two lists do not intersect, so return null.

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.

# 思路 #

首先遍历一遍两条链表，直到其中一个指针到达末尾，这时候有三种情况：

*  a还未到达末尾，说明A链表比B链表长
 *  b还未到达末尾，说明B链表比A链表长
 *  同时到达末尾，说明两根链表一样长

对于情况1，继续将a往后移动同时将A链表的头指针往后移动，当a到达末尾时，从headA算起，A链和B链一样长。此时再同时移动headA和headB，判断指针是否相同，相同即为交叉点并返回

情况2同理  
情况3则直接进行头指针移动

时间复杂度O(n)  
空间复杂度O(1)

## 代码 ##

public class Solution { 
        public ListNode getIntersectionNode(ListNode headA, ListNode headB) { 
        	ListNode a = headA,b = headB;
        	int step = 0;
        	while(a != null && b != null){ 
        		a = a.next;
        		b = b.next;
        	}
            
            while(a != null){ 
            	a = a.next;
            	headA = headA.next;
            }
            
            while(b != null){ 
            	b = b.next;
            	headB = headB.next;
            }
    
            while(headA != null || headB != null){ 
            	if(headA == headB) return headA;
            	else{ 
            		headA = headA.next;
            		headB = headB.next;
            	}
    
            }
            return null;
        }
    }

## 提交结果 ##

Runtime: 1 ms, faster than 98.31% of Java online submissions for Intersection of Two Linked Lists.  
Memory Usage: 34.3 MB, less than 100.00% of Java online submissions for Intersection of Two Linked Lists.

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