*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.

思路 方法超屌！！！

1. 得到2个链条的长度。

2. 将长的链条向前移动差值（len1 - len2）

3. 两个指针一起前进，遇到相同的即是交点，如果没找到，返回null.

相当直观的解法。空间复杂度O(1)， 时间复杂度O(m+n)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        
        ListNode cur = headA;
        int len1 = getLen(headA);
        int len2 = getLen(headB);
        
        int cnt = Math.abs(len1 - len2);
        
        // cut the longer list.
        if (len1 > len2) {
            while (cnt > 0) {
                headA = headA.next;
                cnt--;
            }
        } else {
            while (cnt > 0) {
                headB = headB.next;
                cnt--;
            }
        }
            
        while (headA != null) {
            if (headA == headB) {
                return headA;
            }
            
            headA = headA.next;
            headB = headB.next;
        }
        
        return null;
    }
    
    public int getLen(ListNode head) {
        int cnt = 0;
        while (head != null) {
            head = head.next;
            cnt++;
        }
        
        return cnt;
    }
}

reference： http://www.cnblogs.com/yuzhangcmu/p/4128794.html