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) {
    }
};*/
class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        int l1 = 0; // 记录两个链表的长度，并让链表较长的先走，令两个链表之后的长度相同
        int l2 = 0;
        auto h1 = pHead1;
        auto h2 = pHead2;
        while(h1) {
            h1 = h1->next;
            l1++;
        }
        while(h2) {
            h2 = h2->next;
            l2++;
        }
        int n = l1 - l2;
        if(n > 0)
            while(n--)
                pHead1 = pHead1->next;
        else {
            n = abs(n);
            while(n--)
                pHead2 = pHead2->next;
        }
        while(pHead1 != pHead2) {
            pHead1 = pHead1->next;
            pHead2 = pHead2->next;
        }
        return pHead1;
    }
};