|
| 1 | +- 问题: |
| 2 | + - 给定一个链表的头节点,要求展平一个双向链表。这意味着将链表中的嵌套结构(比如 **`child`** 指针)转换为一个单一的链表。 |
| 3 | + - 例子: |
| 4 | + - 原输入: |
| 5 | + |
| 6 | + ```cpp |
| 7 | + 1 -> 2 -> 3 |
| 8 | + | |
| 9 | + 4 -> 5 |
| 10 | + ``` |
| 11 | + |
| 12 | + - 输出: |
| 13 | + |
| 14 | + ```cpp |
| 15 | + 1 -> 4 -> 5 -> 2 -> 3 |
| 16 | + ``` |
| 17 | + |
| 18 | + - 算法思想:前序遍历 |
| 19 | + |
| 20 | +```cpp |
| 21 | +#include <iostream> |
| 22 | +#include <list> |
| 23 | +using namespace std; |
| 24 | + |
| 25 | +struct Node{ |
| 26 | + Node* next; |
| 27 | + Node* prev; |
| 28 | + Node* child; |
| 29 | + int value; |
| 30 | +}; |
| 31 | + |
| 32 | +void FlattenList(Node* node, std::list<Node>& res){ |
| 33 | + if(node == nullptr){ |
| 34 | + return; |
| 35 | + } |
| 36 | + |
| 37 | + res.push_back(*node); |
| 38 | + FlattenList(node->child, res); |
| 39 | + FlattenList(node->next, res); |
| 40 | +} |
| 41 | + |
| 42 | +std::list<Node> FlattenListHelper(Node* head){ |
| 43 | + std::list<Node> res; |
| 44 | + FlattenList(head, res); |
| 45 | + return res; |
| 46 | +} |
| 47 | + |
| 48 | +int main() { |
| 49 | + // origin list: |
| 50 | + // 1 -> 2 -> 3 |
| 51 | + // | |
| 52 | + // 4 -> 5 |
| 53 | + |
| 54 | + Node* head = new Node{ nullptr, nullptr, nullptr, 1 }; |
| 55 | + head->next = new Node{ nullptr, nullptr, nullptr, 2 }; |
| 56 | + head->next->prev = head; |
| 57 | + head->next->next = new Node{ nullptr, nullptr, nullptr, 3 }; |
| 58 | + head->next->next->prev = head->next; |
| 59 | + head->child = new Node{ nullptr, nullptr, nullptr, 4 }; |
| 60 | + head->child->next = new Node{ nullptr, nullptr, nullptr, 5 }; |
| 61 | + head->child->next->prev = head->child; |
| 62 | + |
| 63 | + std::list<Node> flattenedList = FlattenListHelper(head); |
| 64 | + |
| 65 | + for (const Node& node : flattenedList) { |
| 66 | + cout << node.value << " ";// 1 -> 4 -> 5 -> 2 -> 3 |
| 67 | + } |
| 68 | + |
| 69 | + // Clean up memory |
| 70 | + Node* current = head; |
| 71 | + while (current != nullptr) { |
| 72 | + Node* temp = current; |
| 73 | + current = current->next; |
| 74 | + delete temp; |
| 75 | + } |
| 76 | + |
| 77 | + return 0; |
| 78 | +} |
| 79 | +``` |
0 commit comments