题解 | #两个链表的第一个公共结点#

1.快慢指针法，啥也不是

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        int m = 0,n=0;
        ListNode* temp = pHead1;
        while(temp) {
            m++;
            temp = temp->next;
        }
        temp = pHead2;
        while(temp) {
            n++;
            temp = temp->next;
        }
        if(m>n){
            for(int i=0;i<m-n;i++) {
                pHead1 = pHead1->next;
            }
        } else {
            for(int i=0;i<n-m;i++) {
                pHead2 = pHead2->next;
            }
        }
        while(pHead1!=pHead2 && pHead1 && pHead2) {
            pHead1 = pHead1->next;
            pHead2 = pHead2->next;
        }
        return pHead1;
    }
};