Leecode刷题之旅-C语言/python-203移除链表元素

/*
 * @lc app=leetcode.cn id=203 lang=c
 *
 * [203] 移除链表元素
 *
 * https://leetcode-cn.com/problems/remove-linked-list-elements/description/
 *
 * algorithms
 * Easy (39.58%)
 * Total Accepted:    18.3K
 * Total Submissions: 46.2K
 * Testcase Example:  '[1,2,6,3,4,5,6]\n6'
 *
 * 删除链表中等于给定值 val 的所有节点。
 * 
 * 示例:
 *       p  q
 * 输入: 1->2->6->3->4->5->6, val = 6
 * 输出: 1->2->3->4->5
 * 
 * 
 */
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {
    struct ListNode *p;
    if(head==NULL){
        return 0;
    }
    while(head!=NULL&&head->val==val)
        head = head->next;

    if(head==NULL)
        return 0;

    p = head;
    while(p->next!=NULL){
        if(p->next->val==val)
        p->next=p->next->next;
        else
        {
            p = p->next;
        }
    }
    return head;
}

思路非常简单,相同的删去断链就行。然后开头检查直到head不等于val。

------------------------------------------------------------------------------------------------

python:

#
# @lc app=leetcode.cn id=203 lang=python3
#
# [203] 移除链表元素
#
# https://leetcode-cn.com/problems/remove-linked-list-elements/description/
#
# algorithms
# Easy (39.58%)
# Total Accepted:    18.3K
# Total Submissions: 46.2K
# Testcase Example:  '[1,2,6,3,4,5,6]\n6'
#
# 删除链表中等于给定值 val 的所有节点。
# 
# 示例:
# 
# 输入: 1->2->6->3->4->5->6, val = 6
# 输出: 1->2->3->4->5
# 
# 
#
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        while head is not None and head.val == val:
            head = head.next
        current = head
        while current is not None:
            if current.next is not None and current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next
        return head

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转载自www.cnblogs.com/lixiaoyao123/p/10550854.html