203. Remove Linked List Elements。
2017-09-16 21:11
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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
很基础的题,对链表进行遍历查找删除节点,由于给的节点第一个就有数据,这其实不好计算,可以使用一个存放空数据的空的头结点来简化计算过程。
Example
Given: 1 –> 2 –> 6 –> 3 –> 4 –> 5 –> 6, val = 6
Return: 1 –> 2 –> 3 –> 4 –> 5
很基础的题,对链表进行遍历查找删除节点,由于给的节点第一个就有数据,这其实不好计算,可以使用一个存放空数据的空的头结点来简化计算过程。
#include <iostream> using namespace std; /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* removeElements(ListNode* head, int val) { ListNode temp(0); temp.next = head;//自己创造一个头结点 ListNode* p = &temp; ListNode* q = head; while(q) { if(q->val == val) { p->next = q->next; } else { p = p->next; } q = q->next; } return temp.next; } }; int main() { Solution s; ListNode node1(1); ListNode node2(2); ListNode node3(6); ListNode node4(3); ListNode node5(4); ListNode node6(5); ListNode node7(6); node1.next = &node2; node2.next = &node3; node3.next = &node4; node4.next = &node5; node5.next = &node6; node6.next = &node7; s.removeElements(&node1,6); }
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