3.从尾到头打印链表

题目描述
输入一个链表，按链表从尾到头的顺序返回一个ArrayList。

1)递归
python:

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


# 递归
class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        if listNode is None:
            return []
        else:
            return self.printListFromTailToHead(listNode.next) + [listNode.val]


if __name__ == '__main__':
    s = Solution()
    l = list(map(int, input().strip().split()))
    listNode = ListNode(l[0])
    cur = listNode
    for i in range(1, len(l)):
        cur.next = ListNode(l[i])
        cur = cur.next
    cur.next = None
    print(s.printListFromTailToHead(listNode))

c++:

// 递归
class Solution {
public:
    vector<int> res;
    vector<int> printListFromTailToHead(ListNode* head) {
        if(head != NULL){
            if(head->next != NULL){
                printListFromTailToHead(head->next);
            }
            res.push_back(head->val);
        }
        return res;
    }
};

2)栈的思想

python:

class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        if listNode is None:
            return []
        else:
            stack = []
            res = []
            while listNode is not None:
                stack.append(listNode.val)
                listNode = listNode.next
            while len(stack) > 0:
                res.append(stack.pop())
            return res

c++:

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

3)数组反转
python:

# 数组反转
class Solution:
    # 返回从尾部到头部的列表值序列，例如[1,2,3]
    def printListFromTailToHead(self, listNode):
        if listNode is None:
            return []
        else:
            res = []
            while listNode:
                res.append(listNode.val)
                listNode = listNode.next
            i, j = 0, len(res) - 1
            while i < j:
                res[i], res[j] = res[j], res[i]
                i += 1
                j -= 1
            return res

c++:

//数组反转
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> res;
        while(head != NULL){
            res.push_back(head->val);
            head = head->next;
        }
        int temp = 0;
        int i=0,j=res.size()-1;
        // 数组反转
        while(i<j){
            temp = res[i];
            res[i]=res[j];
            res[j]=temp;
            i++;
            j--;
        }
        return res;
    }
};