找出两链表的交点

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

思路：两个指针，先同时往下，等到第一个到末尾，第二个如果没到末尾，第一个将指针指向第二条头节点，反过来将第二个指针指向第一个头节点，如此，两指针指向到两链表末相等距离的指针

/**
* 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) {
ListNode p1 = headA,p2 = headB;
if(p1==null || p2==null) return null;
while(p1!=p2){
p1 = p1!=null?p1.next:headB;
p2 = p2!=null?p2.next:headA;
}
return p1;
}
}

这里有一点要注意：当两链表无交点时，两指针会最后同时指向null

思路二：用一个函数计算两链表长度，然后算出两起始指针

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
int lenA = length(headA), lenB = length(headB);
// move headA and headB to the same start point
while (lenA > lenB) {
headA = headA.next;
lenA--;
}
while (lenA < lenB) {
headB = headB.next;
lenB--;
}
// find the intersection until end
while (headA != headB) {
headA = headA.next;
headB = headB.next;
}
return headA;
}

private int length(ListNode node) {
int length = 0;
while (node != null) {
node = node.next;
length++;
}
return length;