leetcode160. Intersection of Two Linked Lists

最新推荐文章于 2025-09-11 18:44:42 发布

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原文链接：http://www.cnblogs.com/witchgogogo/p/7274836.html

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#数据结构与算法

本文介绍两种高效查找两个单链表交汇节点的方法：第一种通过计算链表长度并同步遍历来定位；第二种采用双指针法，巧妙地利用指针循环达到同步效果，实现O(n)时间复杂度和O(1)空间复杂度。

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.

解法1:

首先，计算链表A和链表B的长度，其次移动指针至二者剩余长度相同的地方，再两个链表每次移动1个单位判断值是否相同

var getIntersectionNode = function(headA, headB) {
    if(!headA || !headB) {return;}
    var lenA = length(headA),lenB = length(headB);
    while(lenA>lenB) {
        headA = headA.next;
        lenA--;
    }
    while(lenA<lenB) {
        headB = headB.next;
        lenB--;
    }
    while (headA != headB) {
        headA = headA.next;
        headB = headB.next;
    }
    return headA;
};
var length = function(head) {
    var len = 0;
    while(head != null) {
        len++;
        head = head.next;
    }
    return len;
};

解法2:

两链表不同时向后移动一个单位，当短的链表循环完后指向长链表头，因此第二轮循环时指针会指向长度相同的位置

var getIntersectionNode = function(headA, headB) {
    if(headA == null || headB == null) return null;
    var a = headA;
    var b = headB;
    while( a != b){
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }
    return a;
};