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

解题思路

定义两个指针pre和cur，如果cur的值为val，则删除该结点。需要注意的情况有两种：①需要删除头结点；②链表为空。

实现代码[C++]

//Runtime:38ms#include 
     
      using namespace std;struct ListNode{    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL){}};class Solution {public:    ListNode* removeElements(ListNode* head, int val) {        while (head != NULL && head->val == val)        {            head = head->next;        }        if (head == NULL)        {            return NULL;        }        ListNode *pre = head;        ListNode *cur = head->next;        while(cur != NULL)        {            if (cur->val == val)            {                pre->next = pre->next->next;            }            else            {                pre = pre->next;            }            cur = cur->next;        }        return head;    }};void print(ListNode *head){    while (head != NULL)    {        cout<
      
       val;        head = head->next;    }    cout<
       
        next = node2;    node2->next = node3;    node3->next = node4;    print(node1);    node1 = s.removeElements(node1, 2);    print(node1);}
       
      
     

实现代码[Python]

# Rumtime:181msclass ListNode:    def __init__(self, x):        self.val = x        self.next = Noneclass Solution:    # @param {ListNode} head    # @param {integer} val    # @return {ListNode}    def removeElements(self, head, val):        while head != None and head.val == val:            head = head.next        if head == None:            return None        pre = head        cur = head.next        while cur != None:            if cur.val == val:                pre.next = pre.next.next            else:                pre = pre.next            cur = cur.next        return headdef printList(head):    # @param {ListNode} head    while head != None:        print(head.val, end = ' ')        head = head.next    print() # print '\n'node1 = ListNode(1)node2 = ListNode(2)node3 = ListNode(1)node4 = ListNode(2)node1.next = node2node2.next = node3node3.next = node4printList(node1) # print the origin lists = Solution();node1 = s.removeElements(node1, 1)printList(node1) # print the result list