题解 | #从尾到头打印链表#

思路

看到这道题的第一思路就是用栈。

于是我写了下面的Java代码：

/**
*    public class ListNode {
*        int val;
*        ListNode next = null;
*
*        ListNode(int val) {
*            this.val = val;
*        }
*    }
*
*/
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
    public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        Stack<Integer> stack = new Stack<>();
        ArrayList<Integer> list = new ArrayList<>();
        while (listNode != null) {
            stack.push(listNode.val);
            listNode = listNode.next;
        }
        while (!stack.empty()) {
            list.add(stack.pop());
        }
        return list;
    }
}

后来又想起来了之前做过的一道链表反转的题目，其实这道题也是链表反转，于是我写了这样一段代码：

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> ret;
        ListNode* temp = NULL;
        ListNode* newHead = NULL;
        while (head)
        {
            temp = head;
            head = head->next;
            temp->next = newHead;
            newHead = temp;
        }

        while (newHead) 
        {
            ret.push_back(newHead->val);
            newHead = newHead->next;
        }
        return ret;
    }
};

另外还有一种解法就是利用std的reverse：

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> ret;
        while (head)
        {
            ret.push_back(head->val);
            head = head->next;
        }
        std::reverse(ret.begin(), ret.end());
        return ret;
    }
};