203. Remove Linked List Elements

203. Remove Linked List Elements

Remove all elements from a linked list of integers that have value val.

Example Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6 Return: 1 --> 2 --> 3 --> 4 --> 5

Credits: Special thanks to @mithmatt for adding this problem and creating all test cases.

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */public class Solution { public ListNode removeElements(ListNode head, int val) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode p = dummy; ListNode q = head; while(q!=null) { if(q.val == val) { p.next = q.next; } else { p = p.next; } q = q.next; } return dummy.next; } }

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */public class Solution { public ListNode removeElements(ListNode head, int val) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode p = dummy; ListNode q = head; while(q!=null) { if(q.val == val) { p.next = q.next; } else { p = p.next; } q = q.next; } return dummy.next; } }

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */class Solution { public ListNode removeElements(ListNode head, int val) { if (head == null) { return null; } //Dummy head to help us with the 2 pointers ListNode dummyHead = new ListNode(val + 1); dummyHead.next = head; //We keep 2 pointers, previous & current ListNode prev = dummyHead; ListNode curr = prev.next; while (curr != null) { //If we find the value, our previous will point to the current's next //Note that when we find the value we do not move the previous forward //We just move it if we do not find the value in the current node if (curr.val == val) { prev.next = curr.next; } else { prev = curr; } curr = curr.next; } //Then we just return our "official" head return dummyHead.next; }}

/** * 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* tmp = head; head = new ListNode(-1); head->next = tmp; ListNode* walk = head; while (walk) { if (walk->next && walk->next->val == val) { walk->next = walk->next->next; } else { walk = walk->next; } } return head->next; }};

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */public class Solution { public ListNode removeElements(ListNode head, int val) { while(head != null && head.val == val) { head = head.next; } ListNode curr = head; while(curr != null && curr.next != null) { if (curr.next.val == val) { curr.next = curr.next.next; } else { curr = curr.next; } } return head; } }

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