160. 两个链表的相交点 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.

题意：找出两个两个链表的相交点

解法：先算出两个链表的长度，同时遍历两个链表，较短的链表从位于Math.Abs(countA - countB)的节点开始遍历，直到两个指针指向的节点相同

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
		 public ListNode GetIntersectionNode(ListNode headA, ListNode headB) {
			int countA = GetCount(headA);
			int countB = GetCount(headB);
			int offset = Math.Abs(countA - countB);
			if (countA > countB) {
				headA = GetIndex(headA, offset);
			} else {
				headB = GetIndex(headB, offset);
			}
			while (headA != null && headB!=null) {
				if (headA.GetHashCode() == headB.GetHashCode()) {
					return headA;
				} else {
					headA = headA.next;
					headB = headB.next;
				}
			}
			return null;
		}
		 public ListNode GetIndex(ListNode head,int index) {
			int count = 0;
			ListNode node = head;
			while (node != null) {
				if (count == index) {
					return node;
				} else {
					count++;
					node = node.next;
				}				
			}
			return null;
		}
		 public int GetCount(ListNode head) {
			int count = 0;
			ListNode node = head;
			while (node != null) {
				count++;
				node = node.next;
			}
			return count;
		}
}

null