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.

移除所有和给定值相等的链表元素。

解法1：迭代

解法2: 递归

Java:

/**

 * 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) {

        if (head == null) return null;

        head.next = removeElements(head.next, val);

        return head.val == val ? head.next : head;

    }

}　　

Java:

public ListNode removeElements(ListNode head, int val) {

        if(head == null) return head;

        if(head.val == val) return removeElements(head.next, val);

        ListNode preMark = head, nextMark = head;

        while(nextMark.next != null && nextMark.next.val != val){

            nextMark = nextMark.next;

        }

// This line could be deleted, i kept it here for a full logic cycle.

        if(nextMark.next == null) return head;

        preMark = nextMark;

        nextMark = nextMark.next;

        preMark.next = removeElements(nextMark, val);

        return head;

    }　　

Python:

class Solution(object):

    def removeElements(self, head, val):

        """

        :type head: ListNode

        :type val: int

        :rtype: ListNode

        """

        if not head:

            return None

        head.next = self.removeElements(head.next, val)

        return head.next if head.val == val else head　　

Python: wo from G

class Solution(object):

    def removeElements(self, head, val):

        """

        :type head: ListNode

        :type val: int

        :rtype: ListNode

        """

        while head and head.val == val:

            head = head.next

        if head:

            head.next = self.removeElements(head.next, val)

        return head 　

Python:

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    # @param {ListNode} head

    # @param {integer} val

    # @return {ListNode}

    def removeElements(self, head, val):

        dummy = ListNode(float("-inf"))

        dummy.next = head

        prev, curr = dummy, dummy.next

        while curr:

            if curr.val == val:

                prev.next = curr.next

            else:

                prev = curr

            curr = curr.next

        return dummy.next　　

Python:

def removeElements(self, head, val):

        dummy = ListNode(-1)

        dummy.next = head

        pointer = dummy

        while(pointer.next):

            if pointer.next.val == val:

                pointer.next = pointer.next.next

            else:

                pointer = pointer.next

        return dummy.next　　

C++: Iterration

class Solution {

public:

    ListNode* removeElements(ListNode* head, int val) {

        ListNode *dummy = new ListNode(-1), *pre = dummy;

        dummy->next = head;

        while (pre->next) {

            if (pre->next->val == val) {

                ListNode *t = pre->next;

                pre->next = t->next;

                t->next = NULL;

                delete t;

            } else {

                pre = pre->next;

            }

        }

        return dummy->next;

    }

};

C++: Recursion　　

class Solution {

public:

    ListNode* removeElements(ListNode* head, int val) {

        if (!head) return NULL;

        head->next = removeElements(head->next, val);

        return head->val == val ? head->next : head;

    }

};

　　

All LeetCode Questions List 题目汇总

[LeetCode] 203. Remove Linked List Elements 移除链表元素的相关教程结束。