Intersection of Two Linked Lists 找出2个链表的交点

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.

限制：O ( n )的时间复杂度 和 O ( 1 ) 的空间复杂度。

如图链表A 和 B 的长度是不一样的。

要是A 还在a1出发，B在b2出发，大家走的步数一样，就可以一起到达c1了。

一种方法是分别计算链表A 和 链表 B的长度，lenA, lenB ,谁长，谁先出发 abs ( lenB - lenA )步。

运行时间：

代码：

public class IntersectionofTwoLinkedLists {    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        int lenA = listLength(headA);        int lenB = listLength(headB);        ListNode curA = headA, curB = headB;        int diff = lenA - lenB;        while (diff > 0) {            curA = curA.next;            diff--;        }        while (diff < 0) {            curB = curB.next;            diff++;        }        while (curA != null && curB != null) {            if (curA == curB) {                return curA;            }            curA = curA.next;            curB = curB.next;        }        return null;    }    private int listLength(ListNode head) {        int count = 0;        while (head != null) {            head = head.next;            count++;        }        return count;    }}

另外一种是比较巧妙的：

大家都走lenA + lenB步。

代码：

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {    //boundary check    if(headA == null || headB == null) return null;    ListNode a = headA;    ListNode b = headB;    //if a & b have different len, then we will stop the loop after second iteration    while( a != b){        //for the end of first iteration, we just reset the pointer to the head of another linkedlist        a = a == null? headB : a.next;        b = b == null? headA : b.next;        }    return a;}

参考：

https://leetcode.com/discuss/66203/java-solution-without-knowing-the-difference-in-len