LeetCode-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.

我的解决方案是，先计算出两个链表的长度lengthA，lengthB，然后让较长链表的指针先next长度之差次，然后两个指针一块指向，比较地址是否一样，一样返回，否则为null。比如上例，pa、pb分别指向A、B链表的头。因为链表B比链表A的长度长1，所以先让pb下移1次，指向b2的位置，之后两个指针一块下移，判断地址相等。上代码

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode pa = headA;ListNode pb = headB;int lengthA = lengthOfList(headA);int lengthB = lengthOfList(headB);if (lengthA - lengthB > 0) {for (int i = 0; i < lengthA-lengthB; i++) {pa = pa.next;}}if (lengthB - lengthA > 0) {for (int i = 0; i < lengthB-lengthA; i++) {pb = pb.next;};}while (pa != null && pb != null) {if (pa == pb) {return pa;}pa = pa.next;pb = pb.next;}        return null;    }public int lengthOfList(ListNode head) {ListNode p;int length=0;for (p = head; p != null; p = p.next) {length++;}return length;}

还有一种是环形链表性的解法，顺便也贴出来

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode p1 = headA;    ListNode p2 = headB;    if (p1 == null || p2 == null)     return null;    while (p1 != null && p2 != null && p1 != p2) {        p1 = p1.next;        p2 = p2.next;                if (p1 == p2)         return p1;              if (p1 == null)        p1 = headB;        if (p2 == null)         p2 = headA;    }    return p1;    }