first_part = str[:n] # Create a new string 'last_part' that includes all characters from the character at index 'n+1' to the end of 'str'. last_part = str[n+1:] # Return the result by concatenating 'first_part' and 'last_part', effectively removing the character at index 'n'...
Given an integer n, generate the nth sequence. Note: The sequence of integers will be represented as a string. ===Comments by Dabay=== 题意半天没搞懂。原来是给的数字n是几,就返回第几个字符串。例如,如果n是5,就返回“111221”这个字符串。 第一个铁定是1,然后用say的方式来往后生成下一个字...
classSolution:defremoveNthFromEnd(self,head,n):# 初始化一个哑节点,它的下一个节点指向链表头节点# 这样做是为了方便处理边界情况,比如删除的是头节点dummy=ListNode(0)dummy.next=head# 初始化快慢指针,初始时都指向哑节点slow=dummyfast=dummy# 快指针先前进n+1步,走到第n+1个节点# 这里加1是为了让快慢...
等tmp1访问结束的时候,tmp2刚好访问到倒数第n个节点。 代码(python): View Code
在LeetCode新手村100题中，题目19M要求删除链表倒数第N个节点，这是一道中等难度的挑战。解决方法主要有两种，即使用快慢指针和优雅的递归法。快慢指针解法是一种直观且不依赖于链表长度的策略，通过维护两个指针，一个快速移动一个节点，另一个保持固定速度，当快指针到达链表尾部时，慢指针正好在倒数第N...
[Leetcode][python]Remove Nth Node From End of List 题目大意 删除链表中倒数第n个节点 解题思路 加一个虚假头结点dummy,并使用双指针p1和p2。p1先向前移动n个节点(从dummy节点开始移动,所以移动了n其实是移动到了前一位),然后p1和p2同时移动,当p1.next==None时,此时p2.next指的就是需要删除的节点前面一...
Given the head of a linked list, remove the nth node from the end of the list and return its head. The idea is to first find the lengh of the listnode, then count and find the n th node that we want…
python # 0019.删除倒数第N个节点 # 参考:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/solution/dong-hua-tu-jie-leetcode-di-19-hao-wen-ti-shan-chu/ class ListNode: def __init__(self, val): self.val = val ...
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