Leetcode160-Intersection of Two Linked Lists-Easy

找出两个链表的交点

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:

begin to intersect at node c1.

 

Example 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.

注意：交点指的是同一个对象（PointA == PointB）！不是两个Node值相同 （PointA.val == PointB.val）

Example 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:

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.

法一：

如果两个链长度相同，那么同时向后移动指针，对应的一个个比下去就能找到，所以只需要把较长的链表变短即可，使得两个链表长度相同：

定义两个指针，分别指向A，B链表头；

分别遍历链表A, B, 得到A，B的长度，计算长度差值diff；

把两个指针移回链表头部，把较长的链表的指针，先从头向后移动diff；

两个指针，一一比较，每次同时向后移动一步。

/**
 * 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) {
        int alength = 0;
        int blength = 0;
        
　　　　　// 链表A B长度
        ListNode aNode = headA;
        ListNode bNode = headB;
        
        while(aNode != null) {
            alength++;
            aNode = aNode.next;
        }
        
        while(bNode != null) {
            blength++;
            bNode = bNode.next;
        }
        
　　　　 // 将指针移回头结点 
        aNode = headA;
        bNode = headB;
        
　　　　　// 移动较长链表的指针diff步
        int diff = alength - blength;
        if(diff >= 0) {
            while(diff != 0) {
                diff--;
                aNode = aNode.next;
            }
        } else {
            diff = -1 * diff;
            while(diff != 0) {
                diff--;
                bNode = bNode.next;
            }
        }
        
　　　　// find intersection
        while(aNode != bNode) {
            aNode = aNode.next;
            bNode = bNode.next;
        }
        
        return aNode;
    }
}

 

 

法二：

用环的思想来做，我们让两条链表分别从各自的开头开始往后遍历，当其中一条遍历到末尾时，我们跳到另一个条链表的开头继续遍历。两个指针最终会相等，而且只有两种情况，一种情况是在交点处相遇，另一种情况是在各自的末尾的空节点处相等。为什么一定会相等呢，因为两个指针走过的路程相同，是两个链表的长度之和，所以一定会相等。

代码很简洁 

/**
 * 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 aPointer = headA;
        ListNode bPointer = headB;
        
　　　　　// Intersection 条件： aPointer == bPointer, same address! Not aPointer.val == bPointer.val (see example 1)
        while(aPointer != bPointer) {
            aPointer = (aPointer != null) ? aPointer.next : headB;
            bPointer = (bPointer != null) ? bPointer.next : headA;
        }
        
        return aPointer;
    }
}