[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:

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/**  * 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:

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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:

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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

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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:

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# 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:

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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

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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　　

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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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