[LeetCode] 203. Remove Linked List Elements 移除链表元素

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;
    }
};

　　

 

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