考研数据结构：链表基础算法题满分训练

（1）剑指Offer 从尾到头打印链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> printListReversingly(ListNode* head) {
        vector<int> res;
        while(head)
        {
            res.push_back(head->val);
            head = head->next;
        }
        reverse(res.begin(), res.end());
        //return vector<int>(res.rbegin(), res.rend());
        return res;
    }
};

（2）剑指Offer 在O(1)时间删除链表结点（假设链表一定存在，并且该节点一定不是尾节点。） 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void deleteNode(ListNode* node) {
        node->val = node->next->val;
        node->next = node->next->next;
    }
};

（3）剑指Offer 链表中倒数第k个节点 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* findKthToTail(ListNode* head, int k) {
        int n = 0;
        for (ListNode* p = head; p; p = p->next) n ++ ;
        if(k > n) return NULL;
        ListNode* p = head;
        for (int i = 0; i < n - k; i ++ ) p = p->next;
        return p;
    }
};

 （4）剑指Offer, 语法题, Hulu面试题 反转链表

/**
 * 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;
        ListNode* a = head, *b = a->next;
        while(b)
        {
            ListNode *c = b->next;
            b->next = a;
            a = b;
            b = c;
        }
        head->next = NULL;
        return a;
    }
};

（5）剑指Offer 合并两个排序的链表 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1);
        ListNode* tail = dummy;

        while(l1 && l2)
        {
            if(l1->val <= l2->val)
            {
                tail = tail->next = l1;
                l1 = l1->next;
            }
            else
            {
                tail = tail->next = l2;
                l2 = l2->next;
            }
        }
        if(l1) tail->next = l1;
        if(l2) tail->next = l2;
        return dummy->next;
    }
};