[LeetCode] 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

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 <iostream>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<<head->val;head = head->next;}cout<<endl;}int main(){Solution s;ListNode *node1 = new ListNode(1);ListNode *node2 = new ListNode(2);ListNode *node3 = new ListNode(1);ListNode *node4 = new ListNode(2);node1->next = node2;node2->next = node3;node3->next = node4;print(node1);node1 = s.removeElements(node1, 2);print(node1);}实现代码[Python]::self.val = xself.next = ::while head != None and head.val == val:head = head.nextif head == None:return Nonepre = headcur = head.nextwhile cur != None:if cur.val == val:pre.next = pre.next.nextelse:pre = pre.nextcur = cur.nextreturn head:# @param {ListNode} headwhile head != None:print(head.val, end = ‘ ‘)head = head.nextprint() # 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

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