题解 | #两个链表的第一个公共结点#

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
#include <cstdlib>
class Solution {
public:
	int getsize(ListNode* k) {	//获取链表大小
		int count = 0;
		while (k) {
			count++;
			k = k->next;
		}
		return count;
	}
	int abc (int a, int b) {	//绝对值函数
		if (a - b >= 0) {
			return a - b;
		}else{
			return -(a - b);
		}
	}
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
		if (!pHead1 || !pHead2){
			return nullptr;
		}

		int count1 = getsize(pHead1);
		int count2 = getsize(pHead2);
		cout << abc(count1, count2) << endl;
		for (int i = 0; i < abc(count1, count2); i++) {		
		  //哪个链表长，就移动哪个链表，直到两个链表一样长
			if (count1 >= count2) {
				pHead1 = pHead1->next;
			}else{
				pHead2 = pHead2->next;
			}
		}
		while (pHead1 && pHead2) {	//遍历两个链表，当出现相同的值，直接返回这个节点以及后边的节点
			if (pHead1->val == pHead2->val) {
				return pHead1;
			}
			pHead1 = pHead1->next;
			pHead2 = pHead2->next;
		}
		return nullptr;
	}
};