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) {
        ListNode* curr1 = pHead1;
		ListNode* curr2 = pHead2;

		int size_1 = 0;
		int size_2 = 0;
		while (curr1 != nullptr) {
			size_1++;
			curr1 = curr1->next;
		}

		while (curr2 != nullptr) {
			size_2++;
			curr2 = curr2->next;
		}

		curr1 = pHead1;
		curr2 = pHead2;

		if (size_1 < size_2) {
			for (int i = 0; i < size_2 - size_1; i++) {
				curr2 = curr2->next;
			}

			while (curr2 != nullptr)
			{
				if (curr1->val == curr2->val) {
					return curr1;
				} else {
					curr1 = curr1->next;
					curr2 = curr2->next;
				}
			}

		} else {
			for (int i = 0; i < size_1 - size_2; i++) {
				curr1 = curr1->next;
			}
			while (curr1 != nullptr)
			{
				if (curr1->val == curr2->val) {
					return curr2;
				} else {
					curr1 = curr1->next;
					curr2 = curr2->next;
				}
			}
		}


		return nullptr;
    }
};

找到两个链表的长度，然后预处理两个列表到相同长度然后开始遍历对比就可以#剑指offer#