【LeetCode Python實(shí)現(xiàn)】19. 刪除鏈表的倒數(shù)第 N 個(gè)結(jié)點(diǎn)（中等）首次 99.5% +
2022-04-18 10:58:23

---

題目描述

給你一個(gè)鏈表，刪除鏈表的倒數(shù)第 n 個(gè)結(jié)點(diǎn)，并且返回鏈表的頭結(jié)點(diǎn)。

示例 1：

輸入：head = [1,2,3,4,5], n = 2
輸出：[1,2,3,5]

示例 2：

輸入：head = [1], n = 1
輸出：[]

示例 3：

輸入：head = [1,2], n = 1
輸出：[1]

提示：

鏈表中結(jié)點(diǎn)的數(shù)目為 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

參考代碼

第一次遍歷獲取鏈表的長度，第二次遍歷到達(dá)要截取的位數(shù)移除節(jié)點(diǎn)。

#	24 ms	15 MB

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        lens = 1
        temp = head
        while temp.next:
            temp = temp.next
            lens += 1
        
        if lens < n:
            return None
        elif lens == n:
            return head.next
        else:
            temp = head
            for i in range(lens - n - 1):
                temp = temp.next

            if n == 1:
                temp.next = None
            else:
                temp.next = temp.next.next
            
            return head