leetcode160. Intersection of Two Linked Lists

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;
};