題目出處
難度
easy
個人範例程式碼 – two pointers
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
start, end = 0, 0
while(end < len(nums)):
if(nums[end] != 0): # != 0
nums[start], nums[end] = nums[end], nums[start] # change
start +=1
end += 1
return nums
算法說明
同向 pointers,因為題目有說要保留「the relative order」(相對順序)
start 保留最後指向 0 位置 (在指到第一個 0 之前,都是指數字與 end 一起前進,讓自己互換)
最近在練習程式碼本身就可以自解釋的 Coding style,可以嘗試直接閱讀程式碼理解
input handling
處理沒有 input 的狀況,return []
Boundary conditions
當 end 先行達到 len(nums),就結束迴圈
Reference
⭐ Leetcode 解題紀錄 ⭐ | 題型 | 資料結構 | Python Solution | C++ Solution | Note | |
---|---|---|---|---|---|---|
⭐BFS 相關題型 ⭐ | ||||||
104 | Maximum Depth of Binary Tree | BFS (分層) | Python | |||
94 | Binary Tree Inorder Traversal | BFS (分層) | Tree | Python | 內含 處理 Tree 樹問題的重點 | |
102 | Binary Tree Level Order Traversal | BFS (分層) | Tree | Python | ||
103 | Binary Tree Zigzag Level Order Traversal | BFS (分層) | Tree | Python | ||
107 | Binary Tree Level Order Traversal II | BFS (分層) | Tree | Python | ||
133 | Clone Graph | BFS (分層) | Graph | Python | Graph 的基本操作 #重要題型 | |
127 | Word Ladder | BFS (分層), DFS | Graph | Python | ||
[Lint] 127 | Topological Sorting | BFS (拓撲) | Python | 內有 indegree, outdegree 介紹 #重要題型 | ||
207 | Course Schedule | BFS (拓樸) | Graph | Python | ||
210 | Course Schedule II | BFS (拓樸) | Graph | Python | ||
[Lint] 892 | Alien Dictionary | BFS (拓樸) | Graph | Python | ||
[Lint] 431 | Connected Component in Undirected Graph | BFS (連通塊) | Graph | Python | 內含 BFS 模板 #重要題型 | |
1091 | Shortest Path in Binary Matrix | BFS (最短路徑) | Matrix | Python | ||
⭐ Binary Serach 相關題型 ⭐ | ||||||
33 | Search in Rotated Sorted Array | Binary Serach | Array | Python | #重要題型 | |
34 | Find First and Last Position of Element in Sorted Array | Binary Serach | Python | |||
50 | Pow(x, n) | Binary Serach | Python |