LeetCode《图解算法数据结构》链表：图书整理 I

题目

书店店员有一张链表形式的书单，每个节点代表一本书，节点中的值表示书的编号。为更方便整理书架，店员需要将书单倒过来排列，就可以从最后一本书开始整理，逐一将书放回到书架上。请倒序返回这个书单链表。

输入

head = [3,6,4,1]

输出

[1,4,6,3]

解法 1：递归

class Solution {
public:
	std::vector<int> reverseBookList(ListNode* head) {
		recur(head);
		return res;
	}
private:
	std::vector<int> res;
	void recur(ListNode* head) {
		if (head == nullptr) return;
		recur(head->next);
		res.push_back(head->val);
	}
};

解法 2：辅助栈法

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

完整代码

#include <iostream>
#include <vector>
struct ListNode {
	int val;
	ListNode* next;
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
	std::vector<int> reverseBookList(ListNode* head) {
		recur(head);
		return res;
	}
private:
	std::vector<int> res;
	void recur(ListNode* head) {
		if (head == nullptr) return;
		recur(head->next);
		res.push_back(head->val);
	}
};

void test() {
	ListNode* head = new ListNode(1);
	ListNode* n2 = new ListNode(2);
	ListNode* n3 = new ListNode(3);
	ListNode* n4 = new ListNode(4);
	ListNode* n5 = new ListNode(5);
	
	head->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = n5;
	
	Solution solution;
	std::vector<int> reversedList = solution.reverseBookList(head);
	
	std::cout << "Result: " << std::endl;
	for (int num : reversedList) {
		std::cout << num << " ";
	}
	std::cout << std::endl;
	
}

int main() {
	test();
	return 0;
}

#include <iostream>
#include <vector>
#include <stack>
struct ListNode {
	int val;
	ListNode* next;
	ListNode(int x) : val(x), next(NULL) {}
};

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

void test() {
	ListNode* head = new ListNode(1);
	ListNode* n2 = new ListNode(2);
	ListNode* n3 = new ListNode(3);
	ListNode* n4 = new ListNode(4);
	ListNode* n5 = new ListNode(5);
	
	head->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = n5; 
	
	Solution solution;
	std::vector<int> testList = solution.getResultList(head);
	
	std::cout << "Result: " << std::endl;
	for (int x : testList) {
		std::cout << x << " ";
	}
	std::cout << std::endl;
}
int main() {
	test();
	return 0;
}