【LeetCode】Intersection of Two Linked Lists

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

题目要求两个单链表相交的第一个节点，可以通过遍历第一个链表，计算长度len1, 保存最后一个节点，再遍历第二个链表，计算长度len2，同时检查最后一个节点是否相同，之后再从头遍历两链表，（若len1>len2）,链表一先遍历len1-len2个节点，之后同时遍历到相同的节点即可，代码如下：

/**
 * 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) {
        if(headA==null||headB==null){
            return null;
        }
        int len1=0;
        ListNode h1=headA;
        while(h1.next!=null){  //注意是.next
            h1=h1.next;
            len1++;
        }
        int len2=0;
        ListNode h2=headB;
        while(h2.next!=null){
            h2=h2.next;
            len2++;
        }
        ListNode n1=headA;
        ListNode n2=headB;
        if(len1>len2){
            int k=len1-len2;
            while(k-->0){
                n1=n1.next;
            }
        }else{
            int k=len2-len1;
            while(k-->0){
                n2=n2.next;
            }
        }
        /* or
        while(n1!=n2){
			n1=n1.next;
			n2=n2.next;
        }
        return n1;
        */
        while(n1!=null&&n2!=null){
            if(n1==n2){
                return n1;
            }
            n1=n1.next;
            n2=n2.next;
        }
        return null;
    }
}

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