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;
    }