160. 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.

我的解法

/**
 * 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) {
        Stack<ListNode> s1 = new Stack<>();
        Stack<ListNode> s2 = new Stack<>();
        // 链表节点入栈
        while(headA != null){
            s1.push(headA);
            headA = headA.next;
        }
        while(headB != null){
            s2.push(headB);
            headB = headB.next;
        }
        // 两链表前一个交点
        ListNode res = null;
        while(!s1.isEmpty() || !s2.isEmpty()){
            ListNode t1 = (s1.isEmpty()) ? null : s1.pop();
            ListNode t2 = (s2.isEmpty()) ? null : s2.pop();
            // 判断两个链表节点是否为同一对象
            if(t1 != t2)
                return res;
            res = t1;
        }
        return res;
    }
}

算法分析：链表节点入栈，然后倒过来比较。

答案解法（很新颖）

/**
 * 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) {
    // 如果有链表为空，则无交点
    if(headA == null || headB == null) return null;
    
    ListNode a = headA;
    ListNode b = headB;
    
    // 即使a、b长度不同，它们都遍历一遍后，此时节点距离两链表尾节点的距离相等
    while( a != b){
        //第一次循环后，节点变成另一链表的头节点 	
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }
    
    return a;
    }
}

算法分析：将两个链表“拼接”起来，比较“拼起来”的“后半部”链表节点，解决长度不一样导致的无法“倒过来同步比较“的问题。