两个链表的第一个公共结点

两个链表的第一个公共结点

题目描述

输入两个链表，找出它们的第一个公共结点。

思想: 先比较两个链表的长度, 让得出差值, 让长的先走长出的节点数目, 然后两个链表一起走, 有相同节点时, 指向链表的游标会碰在一起, 此时游标就是返回值, 时间复杂度O(m+n)?

class Solution {
public:
    
    unsigned int getLength(ListNode *node) {
        ListNode *ct = node;
        unsigned int count = 0;
        
        while (nullptr != ct) {
            count++;
            ct = ct->next;
        }
        
        return count;
    }
    
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        unsigned int len1 = getLength(pHead1);
        unsigned int len2 = getLength(pHead2);
        
        int len = len1 - len2;
        
        ListNode *ctNode1 = pHead1;
        ListNode *ctNode2 = pHead2;
        
        if (0 < len) {
            for (int i = 0; i < len; i++) {
                ctNode1 = ctNode1->next;
            }
        }
        else {
            for (int i = 0; i < -len; i++) {
                ctNode2 = ctNode2->next;
            }
        }
        
        while ((ctNode1 != ctNode2) && (nullptr != ctNode1) && (nullptr != ctNode2)) {
            ctNode1 = ctNode1->next;
            ctNode2 = ctNode2->next;
        }
        
        return ctNode1;
    }
};

利用栈, 先把两个序列分别压入两个栈中, 然后再依次弹出, 直至不同元素时返回, vs上测试通过, 牛客提交有问题

class Solution {
public:
    ListNode* FindFirstCommonNode(ListNode* pHead1, ListNode* pHead2) {
        stack<ListNode *> sk1;
        stack<ListNode *> sk2;

        ListNode *ct1 = pHead1;
        ListNode *ct2 = pHead2;
        ListNode *ret = nullptr;

        while (nullptr != ct1) {
            sk1.push(ct1);
            ct1 = ct1->next;
        }

        while (nullptr != ct2) {
            sk2.push(ct2);
            ct2 = ct2->next;
        }

        while (sk1.top() == sk2.top()) {
            ret = sk1.top();
            sk1.pop();
            sk2.pop();
        }

        return ret;
    }
};

暴力法, 以此内外层循环, 注意, 第二个while循环后要把游标置于开头位置(我忘了:(), 时间复杂度O(mn)

class Solution {
public:
    ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
        if (nullptr == pHead1 || nullptr == pHead2) {
            return nullptr;
        }
        
        ListNode *ctList1 = pHead1;
        ListNode *ctList2 = pHead2;
        
        while (nullptr != ctList1) {
            while (nullptr != ctList2) {
                if (ctList1->val == ctList2->val)
                    return ctList1;
                ctList2 = ctList2->next;
            }
            ctList2 = pHead2;
            ctList1 = ctList1->next;
        }
        
        return nullptr;
    }
};

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/