每日算法刷题Day14-反转链表、两个链表的第一个公共结点、删除链表中重复的节点

输入:1->2->3->4->5->NULL

输出:5->4->3->2->1->NULL

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head || !head -> next)return head;
        
        auto p = head, q = p -> next;
        
        while(q)
        {
            auto o = q -> next;
            q -> next = p;
            p = q;
            q = o;
        }
        head -> next = NULL;
        return p;
    }
};

给出两个链表如下所示：
A：        a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

输出第一个公共节点c1

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *findFirstCommonNode(ListNode *headA, ListNode *headB) {
        auto p = headA;
        auto q = headB;
        while(p != q)
        {
            if(p) p = p -> next;
            else p = headB;
            if(q) q = q -> next;
            else q = headA;
        
        }
        return p;
    }
};

输入：1->2->3->3->4->4->5

输出：1->2->5

输入：1->1->1->2->3

输出：2->3

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplication(ListNode* head) {
        auto dummy = new ListNode(-1);
        dummy -> next = head;
        auto p = dummy;
        
        while(p -> next)
        {
            auto q = p -> next;
            while(q -> next && p -> next -> val == q -> next -> val)q = q -> next;
            
            if(q == p ->next)p = q;
            else p -> next = q -> next;
        }
        return dummy -> next;
        
    }
};