Posts published in “Graph”

花花酱 LeetCode 797. All Paths From Source to Target

By zxi on March 10, 2018

题目大意：给你一个无环有向图，返回所有从节点0到节点n-1的路径。

Problem:

https://leetcode.com/problems/all-paths-from-source-to-target/description

Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows: the nodes are 0, 1, …, graph.length – 1. graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:

Input: [[1,2], [3], [3], []]

Output: [[0,1,3],[0,2,3]]

Explanation: The graph looks like this:

0--->1

| |

v v

2--->3

There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Note:

The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.

Solution 1: DFS

Time complexity: O(n!)

Space complexity: O(n)

// Author: Huahua

// Running time: 85 ms

class Solution {

public:

vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {

vector<vector<int>> ans;

vector<int> path(1, 0);

dfs(graph, 0, graph.size() - 1, path, ans);

return ans;

}

private:

void dfs(const vector<vector<int>>& graph,

int cur, int dest, vector<int>& path, vector<vector<int>>& ans) {

if (cur == dest) {

ans.push_back(path);

return;

}

for (int next : graph[cur]) {

path.push_back(next);

dfs(graph, next, dest, path, ans);

path.pop_back();

}

};

“Cleaner” Version

class Solution {

public:

vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {

vector<vector<int>> ans;

vector<int> path{0};

dfs(graph, path, ans);

return ans;

}

private:

void dfs(const vector<vector<int>>& graph,

vector<int>& path, vector<vector<int>>& ans) {

if (path.back() == graph.size() - 1) {

ans.push_back(path);

return;

}

for (int next : graph[path.back()]) {

path.push_back(next);

dfs(graph, path, ans);

path.pop_back();

}

};

花花酱 LeetCode 753. Cracking the Safe

By zxi on December 31, 2017

题目大意：让你构建一个最短字符串包含所有可能的密码。

There is a box protected by a password. The password is n digits, where each letter can be one of the first kdigits 0, 1, ..., k-1.

You can keep inputting the password, the password will automatically be matched against the last n digits entered.

For example, assuming the password is "345", I can open it when I type "012345", but I enter a total of 6 digits.

Please return any string of minimum length that is guaranteed to open the box after the entire string is inputted.

Example 1:

Input: n = 1, k = 2

Output: "01"

Note: "10" will be accepted too.

Example 2:

Input: n = 2, k = 2

Output: "00110"

Note: "01100", "10011", "11001" will be accepted too.

Note:

n will be in the range [1, 4].
k will be in the range [1, 10].
k^n will be at most 4096.

Idea: Search

Solution 1: DFS w/ backtracking

C ++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

string crackSafe(int n, int k) {

const int total_len = pow(k, n) + n - 1;

string ans(n, '0');

unordered_set<string> visited{ans};

if (dfs(ans, total_len, n, k, visited))

return ans;

return "";

}

private:

bool dfs(string& ans, const int total_len, const int n, const int k, unordered_set<string>& visited) {

if (ans.length() == total_len)

return true;

string node = ans.substr(ans.length() - n + 1, n - 1);

for (char c = '0'; c < '0' + k; ++c) {

node.push_back(c);

if (!visited.count(node)) {

ans.push_back(c);

visited.insert(node);

if (dfs(ans, total_len, n, k, visited)) return true;

visited.erase(node);

ans.pop_back();

}

node.pop_back();

}

return false;

}

};

Solution 2: DFS w/o backtracking

C++

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

string crackSafe(int n, int k) {

const int total_len = pow(k, n) + n - 1;

string node(n - 1, '0');

string ans;

unordered_set<string> visited;

dfs(node, k, visited, ans);

return ans + node;

}

private:

void dfs(const string& node, const int k, unordered_set<string>& visited, string& ans) {

for (char c = '0'; c < '0' + k; ++c) {

string next = node + c;

if (visited.count(next)) continue;

visited.insert(next);

dfs(next.substr(1), k, visited, ans);

ans.push_back(c);

}

};

花花酱 LeetCode 210. Course Schedule II

By zxi on December 13, 2017

Problem

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

For example:

1	2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1]

1	4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

题目大意：

给你一些课程和它的先修课程，让你输出修课顺序。如果无法修完所有课程，返回空数组。

Idea

Topological sorting

拓扑排序

Solution 1: Topological Sorting

Time complexity: O(V+E)

Space complexity: O(V+E)

C++

// Author: Huahua

// Runtime: 16 ms

class Solution {

public:

vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {

vector<vector<int>> graph(numCourses);

for(const auto& p : prerequisites)

graph[p.second].push_back(p.first);

// states: 0 = unkonwn, 1 == visiting, 2 = visited

vector<int> v(numCourses, 0);

vector<int> ans;

for (int i = 0; i < numCourses; ++i)

if (dfs(i, graph, v, ans)) return {};

std::reverse(ans.begin(), ans.end());

return ans;

}

private:

bool dfs(int cur, vector<vector<int>>& graph, vector<int>& v, vector<int>& ans) {

if (v[cur] == 1) return true;

if (v[cur] == 2) return false;

v[cur] = 1;

for (const int t : graph[cur])

if (dfs(t, graph, v, ans)) return true;

v[cur] = 2;

ans.push_back(cur);

return false;

}

};

Java

// Author: Huahua

// Runtime: 83 ms

class Solution {

public int[] findOrder(int numCourses, int[][] prerequisites) {

ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

for (int i = 0; i < numCourses; ++i)

graph.add(new ArrayList<Integer>());

for (int i = 0; i < prerequisites.length; ++i) {

int course = prerequisites[i][0];

int prerequisite = prerequisites[i][1];

graph.get(course).add(prerequisite);

}

int[] visited = new int[numCourses];

List<Integer> ans = new ArrayList<Integer>();

Integer index = numCourses;

for (int i = 0; i < numCourses; ++i)

if (dfs(i, graph, visited, ans)) return new int[0];

return ans.stream().mapToInt(i->i).toArray();

}

private boolean dfs(int curr, ArrayList<ArrayList<Integer>> graph, int[] visited, List<Integer> ans) {

if (visited[curr] == 1) return true;

if (visited[curr] == 2) return false;

visited[curr] = 1;

for (int next : graph.get(curr))

if (dfs(next, graph, visited, ans)) return true;

visited[curr] = 2;

ans.add(curr);

return false;

}

Posts published in “Graph”

花花酱 LeetCode 797. All Paths From Source to Target

花花酱 LeetCode 753. Cracking the Safe

Idea: Search

Solution 1: DFS w/ backtracking

C ++

Solution 2: DFS w/o backtracking

C++

花花酱 LeetCode 210. Course Schedule II

Problem

Idea

Solution 1: Topological Sorting

C++

Java

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