Leetcode - Intersection of Two Linked Lists

Question

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.

---

Note

• 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.

---

Java Code

public class ListNode {    int val;    ListNode next;    ListNode(int x) {        val = x;        next = null;    }}public ListNode getIntersectionNode(ListNode headA, ListNode headB) {    if(headA == null || headB == null)         return null;    ListNode i = headA;    ListNode j = headB;    int lenA = 1;    int lenB = 1;    //统计两个链表的长度    while((i = i.next) != null) lenA++;    while((j = j.next) != null) lenB++;    //指针i总是指向较长的链表，计算较长链表中需要忽略的前若干个节点    int ignore = 0;    if(lenA >= lenB) {        ignore = lenA - lenB;        i = headA;        j = headB;    }else {        ignore = lenB - lenA;        i = headB;        j = headA;    }    //将节点指针i移到第ignore个节点处    for(int count = 0; count < ignore; ++count)        i = i.next;    //同时遍历两个链表，直到遇到第一个相交的节点    while(i != null){        if(i == j)             return i;        else {            i = i.next;            j = j.next;        }    }    //遍历完所有节点也没找到相交节点    return null;}

说明

• 本题如果没有好的处理办法，那么解决问题会很麻烦，而且复杂度也达不到要求。我们可以通过分析发现，两个有交集的单向链表在相交之后实际上就合并为一条链表了，所以可以找到两者在长度上的等量关系，从而可以轻松地找到两者的第一个交点。