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LintCode | Java Solutions to problems on LintCode/LeetCode | Learning library

 by   awangdev Java Version: Current License: No License

 by   awangdev Java Version: Current License: No License

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kandi X-RAY | LintCode Summary

LintCode is a Java library typically used in Tutorial, Learning, Example Codes, LeetCode applications. LintCode has no bugs, it has no vulnerabilities and it has medium support. However LintCode build file is not available. You can download it from GitHub.
Java Solutions to problems on LintCode/LeetCode
Support
Support
Quality
Quality
Security
Security
License
License
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kandi-support Support

  • LintCode has a medium active ecosystem.
  • It has 4132 star(s) with 1444 fork(s). There are 381 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 0 open issues and 9 have been closed. On average issues are closed in 143 days. There are 1 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of LintCode is current.
LintCode Support
Best in #Learning
Average in #Learning
LintCode Support
Best in #Learning
Average in #Learning

quality kandi Quality

  • LintCode has no bugs reported.
LintCode Quality
Best in #Learning
Average in #Learning
LintCode Quality
Best in #Learning
Average in #Learning

securitySecurity

  • LintCode has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
LintCode Security
Best in #Learning
Average in #Learning
LintCode Security
Best in #Learning
Average in #Learning

license License

  • LintCode does not have a standard license declared.
  • Check the repository for any license declaration and review the terms closely.
  • Without a license, all rights are reserved, and you cannot use the library in your applications.
LintCode License
Best in #Learning
Average in #Learning
LintCode License
Best in #Learning
Average in #Learning

buildReuse

  • LintCode releases are not available. You will need to build from source code and install.
  • LintCode has no build file. You will be need to create the build yourself to build the component from source.
LintCode Reuse
Best in #Learning
Average in #Learning
LintCode Reuse
Best in #Learning
Average in #Learning
Top functions reviewed by kandi - BETA

kandi has reviewed LintCode and discovered the below as its top functions. This is intended to give you an instant insight into LintCode implemented functionality, and help decide if they suit your requirements.

  • Returns the table row for the given file .
  • Returns the closest time in the given time string .
  • Decode a string into a list of lines
  • Search for word in trie .
  • Deletes an item from the heap .
  • Find the parent node .
  • Returns the value of a sorted list .
  • Find by prefix .
  • Check if a word is unique
  • Add an integer value to the map .

LintCode Key Features

Java Solutions to problems on LintCode/LeetCode

Connected Component in Undirected Graph in python

copy iconCopydownload iconDownload
class Solution:
    def connectedSet(self, nodes):
        res = []
        visited = set()

        def dfs(node, path):
            path.append(node.label)
            visited.add(node)
            for neighbor in node.neighbors:
                if neighbor not in visited:
                    dfs(neighbor, path)

        for node in nodes:
            if node not in visited:
                path = []
                dfs(node, path)
                res.append(sorted(path.copy()))
        return res

Python Code Optimization Problem (Lintcode Problem 1886 Moving Targets)

copy iconCopydownload iconDownload
index = [i for i in range(len(nums)) if nums[i] == target]
for i in index:
    nums.insert(0, nums.pop(i))
def MoveTarget(nums, target):
    if len(set(nums)) == 1:
        return nums
    for num in nums:
        if num == target:
            nums.insert(0, nums.pop(num))
-----------------------
index = [i for i in range(len(nums)) if nums[i] == target]
for i in index:
    nums.insert(0, nums.pop(i))
def MoveTarget(nums, target):
    if len(set(nums)) == 1:
        return nums
    for num in nums:
        if num == target:
            nums.insert(0, nums.pop(num))
-----------------------
index = [i for i in range(len(nums)) if nums[i] == target]
for i in index:
    nums.insert(0, nums.pop(i))
def MoveTarget(nums, target):
    if len(set(nums)) == 1:
        return nums
    for num in nums:
        if num == target:
            nums.insert(0, nums.pop(num))
-----------------------
def MoveTarget(nums, target):
    n = nums.count(target)
    nums[:] = [target] * n + [e for e in nums if e != target]

Community Discussions

Trending Discussions on LintCode
  • Connected Component in Undirected Graph in python
  • Python Code Optimization Problem (Lintcode Problem 1886 Moving Targets)
Trending Discussions on LintCode

QUESTION

Connected Component in Undirected Graph in python

Asked 2022-Apr-03 at 15:28

Find connected component in undirected graph. Each node in the graph contains a label and a list of its neighbors. this is the link: https://www.lintcode.com/problem/431/ Leetcode requires membership for this question.

class UndirectedGraphNode:
     def __init__(self, x):
         self.label = x
         self.neighbors = []

I solved it this way but it returns []

class Solution:
    def connectedSet(self, nodes):
        res=[]
        visited=set()
        def dfs(node,path):
            if not node:
                res.append(path.copy())
                return
            if node in visited:
                return
            visited.add(node)
            for neighbor in node.neighbors:              
                if neighbor not in visited:
                    print(path)
                    dfs(neighbor,path+[node.label])
        for node in nodes:
            if node not in visited:
                dfs(node,[])
        return res

This returns []. I belive its logic correct. Run depth first search checking if the node is not visited, till no node exists.

Input Data

 {1,2,4#2,1,4#3,5#4,1,2#5,3}

Expected Result

 [[1,2,4],[3,5]]

Result

[]

ANSWER

Answered 2022-Apr-03 at 04:36

Your condition if not node: doesn't make sense: you're not modifying nodes or making them Falsey. Depth first search explores a forest of rooted trees. Your outer loop over nodes looks for the roots of these trees, so you should just pass a reference to an empty list into dfs, to track all the nodes explored in the search tree for that root:

class Solution:
    def connectedSet(self, nodes):
        res = []
        visited = set()

        def dfs(node, path):
            path.append(node.label)
            visited.add(node)
            for neighbor in node.neighbors:
                if neighbor not in visited:
                    dfs(neighbor, path)

        for node in nodes:
            if node not in visited:
                path = []
                dfs(node, path)
                res.append(sorted(path.copy()))
        return res

Source https://stackoverflow.com/questions/71722858

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install LintCode

You can download it from GitHub.
You can use LintCode like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the LintCode component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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