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41 lines
1.2 KiB
Markdown
41 lines
1.2 KiB
Markdown
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# [203. Remove Linked List Elements](https://leetcode.com/problems/remove-linked-list-elements/description/)
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# 思路
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删除链表中值满足条件的节点,常规操作。
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设两个指针pre和p,p为工作指针,pre为p的前一个节点,判断p的值是否为val:
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* 若是,将pre的next指向p的next,再删除节点p;
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* 若不是,将pre指向p即可。
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最后将p指向pre的next。
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为了操作方便,可以设置一个头结点。
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# C++
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```
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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ListNode* removeElements(ListNode* head, int val) {
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ListNode *list_head = new ListNode(0); // 设一个头结点便于操作
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list_head -> next = head;
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ListNode *pre = list_head, *p = head, *target;
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while(p){
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if(p -> val == val){
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target = p;
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pre -> next = p -> next;
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target -> next = NULL;
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delete target;
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}
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else pre = p;
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p = pre -> next;
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}
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return list_head -> next;
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}
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};
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```
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