#LeetCode 203

This problem let us remove linked list elements.

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.

I will give two ways to solve this problem.

/**
 * 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) {
        //If will be removed element in the first node
        while (head && head -> val == val)
            head = head -> next;  //Update the head node

        ListNode* cur = head;  //Define a pointer,we should keep the head of linked list

        while (head && head -> next)
        {
            //Check the next value of node
            if (head -> next -> val == val)  //If the value==target value
                //Let the current node point to next node of target node
                head -> next = head -> next -> next;
            else head = head -> next;  //Otherwise, update the current node
        }

        return cur;  //return the head of linked list
    }
};

The other way is to set a dummy node.

/**
 * 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) {
      
        ListNode* dummy = new ListNode;  //Create a new node, don't care about its value

        dummy -> next = head;  //Let the new node become the head of linked list

        ListNode* cur = dummy;   //Let a pointer point to the head
  
        //While loop condition: the real head of linked list is not NULL
        while(cur -> next)
        {
            //Check the next node value,if value==target value
            if (cur -> next -> val == val)
                //Change the current node point to the next of target node
                cur -> next = cur -> next -> next;
            else cur = cur -> next;  //Otherwise, update the pointer
        }

        //Return the real head of linked list
        return dummy -> next;
    }
};

Wish the tutorial can help you!