[Offer] to prove safety face questions 05. Replace spaces

topic

Enter a head of the list nodes, each node in turn, the return value (Return an array) from the tail to the head.

Example 1:

输入：head = [1,3,2]
输出：[2,3,1]

limit:
0 <= 链表长度 <= 10000

A thought: Reverse Array

Code

Time complexity: O (n)
complexity of space: O (1)

class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        vector<int> res;
        if (!head) return res;
        while (head != nullptr) {
            res.push_back(head->val);
            head = head->next;
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

Thinking two: Stack

Code

Time complexity: O (n)
space complexity: O (n)

class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        vector<int> res;
        if (!head) return res;
        stack<int> st;
        while (head != nullptr) {
            st.push(head->val);
            head = head->next;
        }
        while (!st.empty()) {
            res.push_back(st.top());
            st.pop();
        }        
        return res;
    }
};