160. 相交链表

题意：

解题思路：

思路：
* [Hash]
* 同时移动两个链表，链表到达尾部，则指向另外一个链表头部，继续遍历，这样会抵消长度差；
* 如果没有相交，则代表长度相等，最后会是 nil == nil，反之有相交；

PHP代码实现：

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val) { $this->val = $val; }
 * }
 */
class Solution {
    /**
     * @param ListNode $headA
     * @param ListNode $headB
     * @return ListNode
     */
    function getIntersectionNode($headA, $headB) {
        if ($headA === null || $headB === null) {
            return null;
        }
        $curA = $headA;
        $curB = $headB;
        while ($curA !== $curB) {
            $curA = ($curA === null) ? $headB : $curA->next;
            $curB = ($curB === null) ? $headA : $curB->next;
        }
        return $curA;
    }
    function getIntersectionNodeHash($headA, $headB) {
        $curA = $headA;
        $curB = $headB;
        $map = [];
        while ($curA != null) {
            array_push($map, $curA);
            $curA = $curA->next;
        }
        while ($curB != null) {
            if (in_array($curB, $map, true)) {
                return $curB;
            }
            $curB = $curB->next;
        }
        return null;
    }
}

GO代码实现：

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
    curA,curB := headA,headB
    for curA != curB {      
        if curA == nil {   
            curA = headB
        } else {
            curA = curA.Next
        }
        if curB == nil {
            curB = headA
        } else {
            curB = curB.Next
        }
    }
    return curA
}