Vzhľadom na prepojený zoznam, kde okrem ďalšie ukazovateľ každý uzol má a dieťa ukazovateľ, ktorý môže alebo nemusí ukazovať na samostatný zoznam. Tieto zoznamy detí môžu mať jeden alebo viac vlastné deti vyrábať a viacúrovňový prepojený zoznam. Vzhľadom na hlavu z prvá úroveň zoznamu. Úlohou je sploštiť zoznam tak, aby sa všetky uzly objavili v a jednoúrovňový prepojený zoznam. Vyrovnajte zoznam tak, aby boli všetky uzly na prvá úroveň by mal prísť najprv druhý úroveň a pod.
Príklady:
Vstup:
výstup: 1->4->6->2->5->7->3->8
Vysvetlenie: Viacúrovňový prepojený zoznam je sploštený, pretože nemá žiadne podradené ukazovatele.
Diskutovali sme sploštenie viacúrovňového prepojeného zoznamu kde uzly majú dva ukazovatele dole a ďalej. V predchádzajúcom príspevku sme sploštený prepojený zoznam úrovni. Ako vyrovnať prepojený zoznam, keď vždy potrebujeme spracovať ukazovateľ nadol pred ďalším v každom uzle.
Obsah
- [Očakávaný prístup] Použitie rekurzie - O(n) čas a O(n) priestor
- [Alternatívny prístup] Použitie zásobníka - O(n) čas a O(n) priestor
C++Prístup je k rekurzívne sploštiť a viacúrovňové prepojené zoznam prechádzaním každého uzla a jeho dcérskych uzlov. Po prvé sploštiť zoznam detí pomocou rekurzie. Once the child list is flattened proceed to the ďalší uzol v poradí. Počas prechodu udržujte a odkaz k predtým navštívený uzol a prepojte ho s aktuálnym uzlom. Tento proces zaisťuje, že všetky uzly z rôznych úrovní sú pripojené do a jeden lineárny zoznam pri zachovaní hĺbkový poriadok.
// A C++ program to flatten a multi- // linked list depth-wise #include using namespace std; class Node { public: int data; Node *next; Node *down; Node(int x) { data = x; next = down = nullptr; } }; void flattenList(Node *curr Node *&prev) { if (curr == nullptr) return; // Add the current element to the list. if (prev != nullptr) prev->next = curr; prev = curr; // Store the next pointer Node *next = curr->next; // Recursively add the bottom list flattenList(curr->down prev); // Recursively add the next list flattenList(next prev); } void printList(Node *head) { Node *curr = head; while (curr != nullptr) { cout << curr->data << ' '; curr = curr->next; } cout << endl; } int main() { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node *head = new Node(5); head->down = new Node(7); head->down->down = new Node(8); head->down->down->down = new Node(30); head->next = new Node(10); head->next->next = new Node(19); head->next->next->down = new Node(22); head->next->next->down->down = new Node(50); head->next->next->next = new Node(28); Node *prev = nullptr; flattenList(head prev); printList(head); return 0; }
Java // A Java program to flatten a multi- // linked list depth-wise class Node { int data; Node next down; Node(int x) { data = x; next = down = null; } } class GfG { static void flattenList(Node curr Node[] prev) { if (curr == null) return; // Add the current element to the list. if (prev[0] != null) prev[0].next = curr; prev[0] = curr; // Store the next pointer Node next = curr.next; // Recursively add the bottom list flattenList(curr.down prev); // Recursively add the next list flattenList(next prev); } static void printList(Node head) { Node curr = head; while (curr != null) { System.out.print(curr.data + ' '); curr = curr.next; } System.out.println(); } public static void main(String[] args) { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); Node[] prev = new Node[1]; flattenList(head prev); printList(head); } }
Python # A Python program to flatten a multi- # linked list depth-wise class Node: def __init__(self x): self.data = x self.next = None self.down = None def flatten_list(curr prev): if curr is None: return # Add the current element to the list. if prev[0] is not None: prev[0].next = curr prev[0] = curr # Store the next pointer next_node = curr.next # Recursively add the bottom list flatten_list(curr.down prev) # Recursively add the next list flatten_list(next_node prev) def print_list(head): curr = head while curr is not None: print(curr.data end=' ') curr = curr.next print() if __name__ == '__main__': # Create a hard coded multi-linked list. # 5 -> 10 -> 19 -> 28 # | | # 7 22 # | | # 8 50 # | # 30 head = Node(5) head.down = Node(7) head.down.down = Node(8) head.down.down.down = Node(30) head.next = Node(10) head.next.next = Node(19) head.next.next.down = Node(22) head.next.next.down.down = Node(50) head.next.next.next = Node(28) prev = [None] flatten_list(head prev) print_list(head)
C# // A C# program to flatten a multi- // linked list depth-wise using System; class Node { public int data; public Node next down; public Node(int x) { data = x; next = down = null; } } class GfG { static void FlattenList(Node curr ref Node prev) { if (curr == null) return; // Add the current element to the list. if (prev != null) prev.next = curr; prev = curr; // Store the next pointer Node next = curr.next; // Recursively add the bottom list FlattenList(curr.down ref prev); // Recursively add the next list FlattenList(next ref prev); } static void PrintList(Node head) { Node curr = head; while (curr != null) { Console.Write(curr.data + ' '); curr = curr.next; } Console.WriteLine(); } static void Main(string[] args) { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); Node prev = null; FlattenList(head ref prev); PrintList(head); } }
JavaScript // A Javascript program to flatten a multi- // linked list depth-wise class Node { constructor(x) { this.data = x; this.next = null; this.down = null; } } function flattenList(curr prev) { if (curr === null) return; // Add the current element to the list. if (prev[0] !== null) prev[0].next = curr; prev[0] = curr; // Store the next pointer let next = curr.next; // Recursively add the bottom list flattenList(curr.down prev); // Recursively add the next list flattenList(next prev); } function printList(head) { let curr = head; while (curr !== null) { console.log(curr.data); curr = curr.next; } } // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 let head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); let prev = [null]; flattenList(head prev); printList(head);
Výstup
5 7 8 30 10 19 22 50 28
[Alternatívny prístup] Použitie zásobníka - O(n) čas a O(n) priestor
C++Prístup je prechádzať viacúrovňový prepojený zoznam pomocou a zásobník . Začnite tým tlačenie na hlavový uzol na zásobník. Potom, zatiaľ čo zásobník nie je prázdny pop horný uzol a spracovať ho. Pre každý uzol TAM jeho ďalší a dole (ak existujú) do zásobníka. Počas tohto procesu prepojenie aktuálneho uzla s predchádzajúcim uzlom udržiavanie zoznamu v sploštenej forme. Priechod zabezpečuje, že uzly zo všetkých úrovní sú spojené v a jednoúrovňový prepojený zoznam zachovanie hĺbkového poriadku.
// A C++ program to flatten a multi- // linked list depth-wise using stack #include using namespace std; class Node { public: int data; Node *next; Node *down; Node(int x) { data = x; next = down = nullptr; } }; void flattenList(Node *head) { if (head == nullptr) return; stack<Node *> st; st.push(head); Node *prev = nullptr; while (!st.empty()) { Node *curr = st.top(); st.pop(); // Push the next node first if (curr->next != nullptr) st.push(curr->next); // Push the bottom node into stack if (curr->down != nullptr) st.push(curr->down); // Add the current element to the list if (prev != nullptr) prev->next = curr; prev = curr; } } void printList(Node *head) { Node *curr = head; while (curr != nullptr) { cout << curr->data << ' '; curr = curr->next; } cout << endl; } int main() { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node *head = new Node(5); head->down = new Node(7); head->down->down = new Node(8); head->down->down->down = new Node(30); head->next = new Node(10); head->next->next = new Node(19); head->next->next->down = new Node(22); head->next->next->down->down = new Node(50); head->next->next->next = new Node(28); flattenList(head); printList(head); return 0; }
Java // A Java program to flatten a multi- // linked list depth-wise using stack import java.util.Stack; class Node { int data; Node next down; Node(int x) { data = x; next = down = null; } } class GfG { static void flattenList(Node head) { if (head == null) return; Stack<Node> stack = new Stack<>(); stack.push(head); Node prev = null; while (!stack.isEmpty()) { Node curr = stack.pop(); // Push the next node first if (curr.next != null) stack.push(curr.next); // Push the bottom node into stack if (curr.down != null) stack.push(curr.down); // Add the current element to the list if (prev != null) prev.next = curr; prev = curr; } } static void printList(Node head) { Node curr = head; while (curr != null) { System.out.print(curr.data + ' '); curr = curr.next; } System.out.println(); } public static void main(String[] args) { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); flattenList(head); printList(head); } }
Python # A Python program to flatten a multi- # linked list depth-wise using stack class Node: def __init__(self x): self.data = x self.next = None self.down = None def flatten_list(head): if head is None: return stack = [head] prev = None while stack: curr = stack.pop() # Push the next node first if curr.next: stack.append(curr.next) # Push the bottom node into stack if curr.down: stack.append(curr.down) # Add the current element to the list if prev: prev.next = curr prev = curr def print_list(head): curr = head while curr: print(curr.data end=' ') curr = curr.next print() if __name__ == '__main__': # Create a hard coded multi-linked list. # 5 -> 10 -> 19 -> 28 # | | # 7 22 # | | # 8 50 # | # 30 head = Node(5) head.down = Node(7) head.down.down = Node(8) head.down.down.down = Node(30) head.next = Node(10) head.next.next = Node(19) head.next.next.down = Node(22) head.next.next.down.down = Node(50) head.next.next.next = Node(28) flatten_list(head) print_list(head)
C# // A C# program to flatten a multi- // linked list depth-wise using stack using System; using System.Collections.Generic; class Node { public int data; public Node next down; public Node(int x) { data = x; next = down = null; } } class GfG { static void FlattenList(Node head) { if (head == null) return; Stack<Node> stack = new Stack<Node>(); stack.Push(head); Node prev = null; while (stack.Count > 0) { Node curr = stack.Pop(); // Push the next node first if (curr.next != null) stack.Push(curr.next); // Push the bottom node into stack if (curr.down != null) stack.Push(curr.down); // Add the current element to the list if (prev != null) prev.next = curr; prev = curr; } } static void PrintList(Node head) { Node curr = head; while (curr != null) { Console.Write(curr.data + ' '); curr = curr.next; } Console.WriteLine(); } static void Main(string[] args) { // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 Node head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); FlattenList(head); PrintList(head); } }
JavaScript // A Javascript program to flatten a multi- // linked list depth-wise using stack class Node { constructor(x) { this.data = x; this.next = null; this.down = null; } } function flattenList(head) { if (head === null) return; let stack = [head]; let prev = null; while (stack.length > 0) { let curr = stack.pop(); // Push the next node first if (curr.next !== null) stack.push(curr.next); // Push the bottom node into stack if (curr.down !== null) stack.push(curr.down); // Add the current element to the list if (prev !== null) prev.next = curr; prev = curr; } } function printList(head) { let curr = head; while (curr !== null) { console.log(curr.data); curr = curr.next; } } // Create a hard coded multi-linked list. // 5 -> 10 -> 19 -> 28 // | | // 7 22 // | | // 8 50 // | // 30 let head = new Node(5); head.down = new Node(7); head.down.down = new Node(8); head.down.down.down = new Node(30); head.next = new Node(10); head.next.next = new Node(19); head.next.next.down = new Node(22); head.next.next.down.down = new Node(50); head.next.next.next = new Node(28); flattenList(head); printList(head);
Výstup
5 7 8 30 10 19 22 50 28
