题目描述

给定一个链表，将其从尾到头打印。

输入：1->2->3->4->5->NULL
返回：[5,4,3,2,1]

• 方法1：反转链表，然后顺序打印

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        head = ReverseList(head);
        vector<int> result;
        ListNode* ptr = head;
        while(ptr != nullptr) {
            result.push_back(ptr->val);
            ptr = ptr->next;
        }
        return result;
    }

    //反转链表
    ListNode* ReverseList(ListNode* head) {
        ListNode* back = head;
        ListNode* pre = nullptr;
        while(back != nullptr && back->next != nullptr) {
            pre = back->next;
            back->next = pre->next;
            pre->next = head;
            head = pre;
        }
        return head;
    }
};

• 方法2：顺序遍历链表，并将值保存到链表中。然后再依次从栈中弹出。

class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        stack<int> stk;
        vector<int> result;
        ListNode* ptr = head;
        while(ptr != nullptr) {
            stk.push(ptr->val);
            ptr = ptr->next;
        }
        while(!stk.empty()) {
            result.push_back(stk.top());
            stk.pop();
        }
        return result;
    }
};

• 方法3：递归

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> result;
        func(head, result);
        return result;
    }

    void func(ListNode* head, vector<int>& result) {
        if(head == nullptr)
            return;
        func(head->next, result);
        result.push_back(head->val);
    }
};