链表相交（C++）

leetcode链接
思路：得到两个链表总长度，并计算它们的差值diff，利用差值让两个链表尾端对齐，即移动长链表的指针，共移动diff步。对齐之后可以同时移动两个链表的指针，两个指针若不相等，则继续移动指针，否则相等并返回指针。
代码：

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {	//链表相交
public:
	ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
		ListNode* curA = headA;
		ListNode* curB = headB;
		int lenA = 0, lenB = 0, diff;
		while (curA != NULL) {
			lenA++;
			curA = curA->next;
		}
		while (curB != NULL) {
			lenB++;
			curB = curB->next;
		}
		diff = abs(lenA - lenB);
		curA = headA;
		curB = headB;
		if (lenA > lenB) {
			while (diff--) {
				curA = curA->next;
			}
		}
		else {
			while (diff--) {
				curB = curB->next;
			}
		}
		while (curA != curB) {
			curA = curA->next;
			curB = curB->next;
		}
		return curA;
	}
};