【2024-10-31-2024-11-03】LeetCode刷题——python语法基础题

用户11029137

发布于 2024-11-04 08:11:13

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代码可运行

文章被收录于专栏：编程学习编程学习

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代码可运行

一，交替合并字符串

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        i, n, m = 0, len(word1), len(word2)
        ans = []
        while i < n or i < m:
            if i < n:
                ans.append(word1[i])
            if i < m:
                ans.append(word2[i])
            i+=1
        return "".join(ans)

⭐笔记：利用一个i同时控制两个字符串

二，移动零

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        l, r =0, 0
        while r < len(nums):
            if l == r or nums[r] == 0:
                r += 1
            elif nums[l] != 0:
                l += 1
            else:
                nums[l] = nums[r]
                nums[r] = 0
        return nums

三，罗马数字转整数

class Solution:
    def romanToInt(self, s: str) -> int:
        ROMAN = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
        ans = 0
        for x, y in pairwise(s):
            x, y = ROMAN[x], ROMAN[y]
            ans += x if x >= y else -x
        return ans + ROMAN[s[-1]]

pairwise 来源于itertools模块，用于遍历s序列中的元素对：将输入的可迭代对象（比如列表、元组等）中的元素按顺序两两配对，然后返回这些配对组成的迭代器。

from itertools import pairwise  
  
s = [1, 2, 3, 4, 5]  
for x, y in pairwise(s):  
    print(x, y)

# 输出：
#  1 2  
#  2 3  
#  3 4  
#  4 5

四，反转链表

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        pre = None
        cur = head
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre

运用类似“头插法的方式”，重点是要记录链表不同时刻（变换后）首元素的位置