✡ leetcode 160. Intersection of Two Linked Lists 求两个链表的起始重复位置 --------- java

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.

如果两个链表有重复的部分，那么返回重复的起始位置，否则，返回null。

两种方法：

1、（参考discuss中，并非最佳答案）

可以利用hashMap，先把第一个链表的所有节点放入map中，然后再遍历第二个链表，看map中是否有相同的节点。如果没有，返回null。

时间复杂度O(m+n),空间复杂度O(m) or O(n)

2、先遍历一次链表，找出连个链表的最后一个节点（如果不一样，那么返回null）以及长度差（最佳）

然后较长的链表先走长度差个节点，

之后两个链表一起走，遇到相同的就返回。

时间复杂度O(m+n),空间复杂度O(1)

/**
 * 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 len1 = 1,len2 = 1;
        if( headA == null || headB == null)
            return null;
        ListNode node1 = headA;
        ListNode node2 = headB;

        while( node1.next != null ){
            node1 = node1.next;
            len1++;
        }
        while( node2.next != null ){
            node2 = node2.next;
            len2++;
        }
        if( len1 == 0 || len2 == 0 || node1 != node2 )
            return null;
        node1 = headA;
        node2 = headB;
        if( len1 > len2 ){
            while( len1 > len2 ){
                len1--;
                node1 = node1.next;
            }
        }else if( len1 < len2 ){
            while( len1<len2){
                len2--;
                node2 = node2.next;
            }
        }

        while( len1 >0 ){
            if( node1 == node2 )
                return node1;
            node1 = node1.next;
            node2 = node2.next;
            len1--;
        }
        return null;
    }
}

3、（参考discuss）

不用第一次遍历找出长度差，一直循环遍历找出相同的节点（如果不存在会出现null）

/**
 * 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) {
    //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;
}
}