题解 | #两个链表的第一个公共结点#

这是2021年的第一题，主要是找到链表长度这个工具，再根据双链表Y的形状，配合快慢指针，找到突破口。tt

/* struct ListNode { int val; struct ListNode next; ListNode(int x) : val(x), next(NULL) { } };/ //null表示符号 class Solution { public: ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) { int L1 = GetLength(pHead1); int L2 = GetLength(pHead2);

		//move bigger 
		int step=0;
		ListNode *fast = NULL, *slow =NULL;
		if(L2 >=L1)
		{
			step = L2 - L1;
			fast=pHead2;
			slow=pHead1;

		}else{
			step = L1 - L2;
			fast=pHead1;
			slow=pHead2;
		
		}
		
		//等长比较
			while(step-- >0)
			{
				fast=fast->next;
			}
			while(fast!=slow && fast!=NULL && slow!=NULL ){
				fast=fast->next;
				slow=slow->next;
			}
			if(fast==NULL){
			return fast;}
			else{
				return fast;
			}
}


    int GetLength(ListNode* pH)
	{
		
		if(pH==NULL )return 0;
		
		int len=0;

		while(pH!=NULL){
			len++;
			pH=pH->next;
		}
        return len;

	
    }

};