{"id":76659,"date":"2024-09-13T22:05:37","date_gmt":"2024-09-13T18:35:37","guid":{"rendered":"https:\/\/nabfollower.com\/blog\/linked-list-manipulation-coding-interview-pattern-15f1\/"},"modified":"2024-09-13T22:05:37","modified_gmt":"2024-09-13T18:35:37","slug":"linked-list-manipulation-coding-interview-pattern-15f1","status":"publish","type":"post","link":"https:\/\/nabfollower.com\/blog\/linked-list-manipulation-coding-interview-pattern-15f1\/","title":{"rendered":"\u062f\u0633\u062a\u06a9\u0627\u0631\u06cc \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc\u060c \u0627\u0644\u06af\u0648\u06cc \u0645\u0635\u0627\u062d\u0628\u0647 \u06a9\u062f\u06af\u0630\u0627\u0631\u06cc"},"content":{"rendered":"

Summarize this content to 400 words in Persian Lang <\/p>\n

\u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f \u0634\u062f\u0647<\/p>\n

\u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc \u06cc\u06a9 \u0633\u0627\u062e\u062a\u0627\u0631 \u062f\u0627\u062f\u0647 \u062e\u0637\u06cc \u0645\u062a\u0634\u06a9\u0644 \u0627\u0632 \u06af\u0631\u0647 \u0647\u0627\u06cc\u06cc \u0627\u0633\u062a \u06a9\u0647 \u062f\u0631 \u0622\u0646 \u0647\u0631 \u06af\u0631\u0647 \u062d\u0627\u0648\u06cc \u062f\u0627\u062f\u0647 \u0648 \u06cc\u06a9 \u0645\u0631\u062c\u0639 (\u06cc\u0627 \u067e\u06cc\u0648\u0646\u062f) \u0628\u0647 \u06af\u0631\u0647 \u0628\u0639\u062f\u06cc \u062f\u0631 \u062f\u0646\u0628\u0627\u0644\u0647 \u0627\u0633\u062a. \u0628\u0631\u062e\u0644\u0627\u0641 \u0622\u0631\u0627\u06cc\u0647 \u0647\u0627\u060c \u0644\u06cc\u0633\u062a \u0647\u0627\u06cc \u067e\u06cc\u0648\u0646\u062f\u06cc \u067e\u0648\u06cc\u0627 \u0647\u0633\u062a\u0646\u062f \u0648 \u0628\u0647 \u0631\u0627\u062d\u062a\u06cc \u0645\u06cc \u062a\u0648\u0627\u0646\u0646\u062f \u0628\u0627 \u0627\u0641\u0632\u0648\u062f\u0646 \u06cc\u0627 \u062d\u0630\u0641 \u0639\u0646\u0627\u0635\u0631 \u0628\u062f\u0648\u0646 \u0646\u06cc\u0627\u0632 \u0628\u0647 \u062c\u0627\u0628\u062c\u0627\u06cc\u06cc \u0639\u0646\u0627\u0635\u0631 \u062f\u06cc\u06af\u0631\u060c \u0628\u0632\u0631\u06af \u06cc\u0627 \u06a9\u0648\u0686\u06a9 \u0634\u0648\u0646\u062f. \u0627\u06cc\u0646 \u0648\u06cc\u0698\u06af\u06cc \u0644\u06cc\u0633\u062a\u200c\u0647\u0627\u06cc \u067e\u06cc\u0648\u0646\u062f\u06cc \u0631\u0627 \u0628\u0631\u0627\u06cc \u067e\u06cc\u0627\u062f\u0647\u200c\u0633\u0627\u0632\u06cc \u0633\u0627\u062e\u062a\u0627\u0631\u0647\u0627\u06cc \u062f\u0627\u062f\u0647\u200c\u0627\u06cc \u0645\u0627\u0646\u0646\u062f \u067e\u0634\u062a\u0647\u200c\u0647\u0627\u060c \u0635\u0641\u200c\u0647\u0627 \u0648 \u0645\u0648\u0627\u0631\u062f \u062f\u06cc\u06af\u0631 \u06a9\u0647 \u0627\u0646\u062f\u0627\u0632\u0647 \u0622\u0646\u200c\u0647\u0627 \u0627\u063a\u0644\u0628 \u062a\u063a\u06cc\u06cc\u0631 \u0645\u06cc\u200c\u06a9\u0646\u062f \u0645\u0641\u06cc\u062f \u0645\u06cc\u200c\u0633\u0627\u0632\u062f.<\/p>\n

\u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f \u0645\u0639\u06a9\u0648\u0633<\/p>\n

This question is part of Leetcode problems, question no. 206.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Reverse Linked List” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* reverseList(ListNode* head) {
\n ListNode* prev = nullptr; \/\/ Initialize previous pointer as nullptr
\n ListNode* current = head; \/\/ Initialize current pointer to head<\/p>\n

while (current != nullptr) {
\n ListNode* nextNode = current->next; \/\/ Store next node
\n current->next = prev; \/\/ Reverse the current node’s pointer
\n prev = current; \/\/ Move prev pointer one step forward
\n current = nextNode; \/\/ Move current pointer one step forward
\n }<\/p>\n

return prev; \/\/ Return the new head of the reversed list
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f \u0645\u0639\u06a9\u0648\u0633 II<\/p>\n

This question is part of Leetcode problems, question no. 92.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Reverse Linked List II” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* reverseBetween(ListNode* head, int left, int right) {
\n if (!head || left == right) return head;<\/p>\n

ListNode* dummy = new ListNode(0); \/\/ Dummy node to simplify edge cases
\n dummy->next = head;
\n ListNode* prev = dummy;<\/p>\n

\/\/ Move `prev` pointer to just before the `left` position
\n for (int i = 1; i left; ++i) {
\n prev = prev->next;
\n }<\/p>\n

ListNode* current = prev->next; \/\/ `current` starts at `left`
\n ListNode* nextNode;<\/p>\n

\/\/ Reverse the sublist between `left` and `right`
\n for (int i = 0; i right – left; ++i) {
\n nextNode = current->next;
\n current->next = nextNode->next;
\n nextNode->next = prev->next;
\n prev->next = nextNode;
\n }<\/p>\n

return dummy->next; \/\/ Return the updated list starting from the original head
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u06af\u0631\u0647 \u0646\u0647\u0645 \u0631\u0627 \u0627\u0632 \u0627\u0646\u062a\u0647\u0627\u06cc \u0644\u06cc\u0633\u062a \u062d\u0630\u0641 \u06a9\u0646\u06cc\u062f<\/p>\n

This question is part of Leetcode problems, question no. 19.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “\u062d\u0630\u0641 \u06af\u0631\u0647 \u0646\u0647\u0645 \u0627\u0632 \u0627\u0646\u062a\u0647\u0627\u06cc \u0644\u06cc\u0633\u062a” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* removeNthFromEnd(ListNode* head, int n) {
\n ListNode* dummy = new ListNode(0);
\n dummy->next = head;
\n ListNode* first = dummy;
\n ListNode* second = dummy;<\/p>\n

for (int i = 1; i n + 1; ++i) {
\n first = first->next;
\n }<\/p>\n

while (first != nullptr) {
\n first = first->next;
\n second = second->next;
\n }<\/p>\n

second->next = second->next->next;
\n return dummy->next;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0686\u0631\u062e\u0647 \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc<\/p>\n

This question is part of Leetcode problems, question no. 141.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 \u0631\u0627\u0647 \u062d\u0644 \u0628\u0631\u0627\u06cc \u0645\u0633\u0626\u0644\u0647 “\u0686\u0631\u062e\u0647 \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n bool hasCycle(ListNode *head) {
\n if (!head || !head->next) return false;<\/p>\n

ListNode* slow = head;
\n ListNode* fast = head->next;<\/p>\n

while (slow != fast) {
\n if (!fast || !fast->next) return false;
\n slow = slow->next;
\n fast = fast->next->next;
\n }<\/p>\n

return true;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0686\u0631\u062e\u0647 \u0641\u0647\u0631\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc II<\/p>\n

This question is part of Leetcode problems, question no. 142.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “List Linked Cycle II” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode *detectCycle(ListNode *head) {
\n if (!head || !head->next) return nullptr;<\/p>\n

ListNode* slow = head;
\n ListNode* fast = head;<\/p>\n

\/\/ Phase 1: Detect if there is a cycle
\n while (fast && fast->next) {
\n slow = slow->next;
\n fast = fast->next->next;
\n if (slow == fast) {
\n break;
\n }
\n }<\/p>\n

\/\/ No cycle detected
\n if (fast == nullptr || fast->next == nullptr) {
\n return nullptr;
\n }<\/p>\n

\/\/ Phase 2: Find the start node of the cycle
\n slow = head;
\n while (slow != fast) {
\n slow = slow->next;
\n fast = fast->next;
\n }<\/p>\n

return slow; \/\/ Both pointers meet at the start of the cycle
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u062a\u0642\u0627\u0637\u0639 \u062f\u0648 \u0644\u06cc\u0633\u062a \u0645\u0631\u062a\u0628\u0637<\/p>\n

This question is part of Leetcode problems, question no. 160.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “\u062a\u0642\u0627\u0637\u0639 \u062f\u0648 \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
\n if (!headA || !headB) return nullptr;<\/p>\n

ListNode* ptrA = headA;
\n ListNode* ptrB = headB;<\/p>\n

\/\/ Traverse both lists. When a pointer reaches the end, redirect it to the head of the other list.
\n \/\/ If they intersect, they will meet at the intersection node. If not, they will end up at the end of the lists together.
\n while (ptrA != ptrB) {
\n ptrA = ptrA ? ptrA->next : headB;
\n ptrB = ptrB ? ptrB->next : headA;
\n }<\/p>\n

return ptrA; \/\/ Either the intersection node or nullptr if no intersection
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u062f\u0648 \u0639\u062f\u062f \u0627\u0636\u0627\u0641\u0647 \u06a9\u0646\u06cc\u062f<\/p>\n

This question is part of Leetcode problems, question no. 2.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “\u0627\u0641\u0632\u0648\u062f\u0646 \u062f\u0648 \u0639\u062f\u062f” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
\n ListNode* dummy = new ListNode(0);
\n ListNode* current = dummy;
\n int carry = 0;<\/p>\n

while (l1 || l2 || carry) {
\n int sum = carry;<\/p>\n

if (l1) {
\n sum += l1->val;
\n l1 = l1->next;
\n }<\/p>\n

if (l2) {
\n sum += l2->val;
\n l2 = l2->next;
\n }<\/p>\n

carry = sum \/ 10;
\n current->next = new ListNode(sum % 10);
\n current = current->next;
\n }<\/p>\n

return dummy->next;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0636\u0627\u0641\u0647 \u06a9\u0631\u062f\u0646 \u062f\u0648 \u0639\u062f\u062f II<\/p>\n

This question is part of Leetcode problems, question no. 445.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0633\u0626\u0644\u0647 “Add Two Numbers II” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
\n stackint> s1, s2;<\/p>\n

\/\/ Push all values of l1 to stack s1
\n while (l1) {
\n s1.push(l1->val);
\n l1 = l1->next;
\n }<\/p>\n

\/\/ Push all values of l2 to stack s2
\n while (l2) {
\n s2.push(l2->val);
\n l2 = l2->next;
\n }<\/p>\n

ListNode* prev = nullptr;
\n int carry = 0;<\/p>\n

\/\/ Process the stacks and build the result list
\n while (!s1.empty() || !s2.empty() || carry) {
\n int sum = carry;<\/p>\n

if (!s1.empty()) {
\n sum += s1.top();
\n s1.pop();
\n }<\/p>\n

if (!s2.empty()) {
\n sum += s2.top();
\n s2.pop();
\n }<\/p>\n

carry = sum \/ 10;
\n ListNode* node = new ListNode(sum % 10);
\n node->next = prev;
\n prev = node;
\n }<\/p>\n

return prev;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u06af\u0631\u0647 \u0647\u0627 \u0631\u0627 \u0628\u0647 \u0635\u0648\u0631\u062a \u062c\u0641\u062a \u0639\u0648\u0636 \u06a9\u0646\u06cc\u062f<\/p>\n

This question is part of Leetcode problems, question no. 24.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “\u062a\u0628\u0627\u062f\u0644 \u06af\u0631\u0647 \u0647\u0627 \u0628\u0647 \u0635\u0648\u0631\u062a \u062c\u0641\u062a” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* swapPairs(ListNode* head) {
\n ListNode* dummy = new ListNode(0);
\n dummy->next = head;
\n ListNode* prev = dummy;<\/p>\n

while (prev->next && prev->next->next) {
\n ListNode* first = prev->next;
\n ListNode* second = prev->next->next;<\/p>\n

\/\/ Swap the nodes
\n first->next = second->next;
\n second->next = first;
\n prev->next = second;<\/p>\n

\/\/ Move prev pointer two nodes ahead
\n prev = first;
\n }<\/p>\n

return dummy->next;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0639\u0646\u0627\u0635\u0631 \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc \u0631\u0627 \u062d\u0630\u0641 \u06a9\u0646\u06cc\u062f<\/p>\n

This question is part of Leetcode problems, question no. 203.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Remove Linked List Elements” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* removeElements(ListNode* head, int val) {
\n \/\/ Remove leading nodes with the value ‘val’
\n while (head != nullptr && head->val == val) {
\n head = head->next;
\n }<\/p>\n

\/\/ If the list becomes empty, return nullptr
\n if (head == nullptr) return head;<\/p>\n

\/\/ Initialize current pointer
\n ListNode* current = head;<\/p>\n

\/\/ Traverse the list and remove nodes with the value ‘val’
\n while (current->next != nullptr) {
\n if (current->next->val == val) {
\n current->next = current->next->next;
\n } else {
\n current = current->next;
\n }
\n }<\/p>\n

return head;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u062f\u063a\u0627\u0645 \u062f\u0648 \u0644\u06cc\u0633\u062a \u0645\u0631\u062a\u0628 \u0634\u062f\u0647<\/p>\n

This question is part of Leetcode problems, question no. 21.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Merge Two Sorted List” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
\n \/\/ Create a dummy node to simplify the merging process
\n ListNode* dummy = new ListNode(0);
\n ListNode* current = dummy;<\/p>\n

\/\/ Merge the two lists while both have remaining nodes
\n while (l1 != nullptr && l2 != nullptr) {
\n if (l1->val l2->val) {
\n current->next = l1;
\n l1 = l1->next;
\n } else {
\n current->next = l2;
\n l2 = l2->next;
\n }
\n current = current->next;
\n }<\/p>\n

\/\/ Append any remaining nodes from either list
\n if (l1 != nullptr) {
\n current->next = l1;
\n } else {
\n current->next = l2;
\n }<\/p>\n

\/\/ Return the merged list starting from dummy->next
\n return dummy->next;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f \u062e\u0648\u0631\u062f\u0647 \u0632\u0648\u062c \u0632\u0648\u062c<\/p>\n

This question is part of Leetcode problems, question no. 328.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Odd Even Linked List” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* oddEvenList(ListNode* head) {
\n if (head == nullptr) return nullptr;<\/p>\n

\/\/ Initialize pointers for odd and even nodes
\n ListNode* odd = head;
\n ListNode* even = head->next;
\n ListNode* evenHead = even; \/\/ Keep track of the start of even list<\/p>\n

\/\/ Iterate through the list to rearrange the nodes
\n while (even != nullptr && even->next != nullptr) {
\n odd->next = even->next; \/\/ Link the odd nodes
\n odd = odd->next; \/\/ Move to the next odd node
\n even->next = odd->next; \/\/ Link the even nodes
\n even = even->next; \/\/ Move to the next even node
\n }<\/p>\n

\/\/ Attach the even list at the end of the odd list
\n odd->next = evenHead;<\/p>\n

return head;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0641\u0647\u0631\u0633\u062a \u067e\u06cc\u0648\u0646\u062f\u06cc \u067e\u0627\u0644\u06cc\u0646\u062f\u0631\u0648\u0645<\/p>\n

This question is part of Leetcode problems, question no. 234.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Palindrome Linked List” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n bool isPalindrome(ListNode* head) {
\n if (!head || !head->next) return true;<\/p>\n

\/\/ Step 1: Find the middle of the linked list using the fast and slow pointer technique
\n ListNode* slow = head;
\n ListNode* fast = head;<\/p>\n

while (fast && fast->next) {
\n slow = slow->next;
\n fast = fast->next->next;
\n }<\/p>\n

\/\/ Step 2: Reverse the second half of the list
\n ListNode* prev = nullptr;
\n while (slow) {
\n ListNode* nextNode = slow->next;
\n slow->next = prev;
\n prev = slow;
\n slow = nextNode;
\n }<\/p>\n

\/\/ Step 3: Compare the first half and the reversed second half
\n ListNode* firstHalf = head;
\n ListNode* secondHalf = prev;<\/p>\n

while (secondHalf) {
\n if (firstHalf->val != secondHalf->val) {
\n return false;
\n }
\n firstHalf = firstHalf->next;
\n secondHalf = secondHalf->next;
\n }<\/p>\n

return true;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\u0648\u0633\u0637 \u0644\u06cc\u0633\u062a \u067e\u06cc\u0648\u0646\u062f \u0634\u062f\u0647<\/p>\n

This question is part of Leetcode problems, question no. 876.<\/p>\n

\u062f\u0631 \u0627\u06cc\u0646\u062c\u0627 \u06a9\u0644\u0627\u0633 Solution \u0628\u0631\u0627\u06cc \u0645\u0634\u06a9\u0644 “Middle of the Linked List” \u062f\u0631 C++ \u0622\u0645\u062f\u0647 \u0627\u0633\u062a:<\/p>\n

class Solution {
\npublic:
\n ListNode* middleNode(ListNode* head) {
\n ListNode* slow = head;
\n ListNode* fast = head;<\/p>\n

\/\/ Move slow by one step and fast by two steps
\n \/\/ When fast reaches the end, slow will be at the middle
\n while (fast && fast->next) {
\n slow = slow->next;
\n fast = fast->next->next;
\n }<\/p>\n

return slow;
\n }
\n};<\/p>\n

\u0648\u0627\u0631\u062f \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u0634\u0648\u06cc\u062f<\/p>\n

\u0627\u0632 \u062d\u0627\u0644\u062a \u062a\u0645\u0627\u0645 \u0635\u0641\u062d\u0647 \u062e\u0627\u0631\u062c \u0634\u0648\u06cc\u062f<\/p>\n

\n
\n
\n

\u0641\u0647\u0631\u0633\u062a \u0645\u0637\u0627\u0644\u0628<\/p>\nToggle<\/span><\/path><\/svg><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n