🗓️Day 03 – Linked list.
Topics: Linked List
Problem solved:
#203 - Remove Linked List Elements
#206 - Reverse Linked List

📚Concept Review – What is linked list?

A linked list is a linear data strcture that made up of nodes. Each nodes contains some infomation, such as value, pointer.
Highlight

• Dynamic size
• Don’t need contiguous memory
• Can’t access randomly
• Easy for insert or delete

---

🧩Problem solution – 203. Remove Linked List Elements

link

💡Problem description

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

Example 1:
Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:
Input: head = [], val = 1
Output: []

Example 3:
Input: head = [7,7,7,7], val = 7
Output: []

Constraints:

The number of nodes in the list is in the range [0, 10 ⁴].
1 <= Node.val <= 50
0 <= val <= 50

🧭Thought Process

If the head need to be removed, then we just move the pointer head to the next one. Besides, it is just cur->next = next->next.

💻Code Implementation

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        while (head != NULL) {
            if (head->val == val) {
                head = head->next;
            } else {
                break;
            }
        }
        if (head == NULL)
            return head;
        ListNode* h = head;
        ListNode* n = head->next;
        while (n != NULL) {
            if (n->val == val) {
                h->next = n->next;
                n = h->next;
            } else {
                h = h->next;
                n = h->next;
            }
        }
        return head;
    }
};

⏱️Complexity Analysis

• Time complexity: O(n)
• Space complexity: O(1)

🧩Problem solution – 206. Reverse Linked List

link

💡Problem description

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:
Input: head = []
Output: []

Constraints:

The number of nodes in the list is the range [0, 5000].
-5000 <= Node.val <= 5000

🧭Thought Process

💻Code Implementation

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* pre = NULL;
        ListNode* cur = head;
        ListNode* temp;
        while(cur!=NULL){
            temp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
};

⏱️Complexity Analysis

• Time complexity: O(n)
• Space complexity: O(1)