# Topological Sorting
### Source
- lintcode: [(127) Topological Sorting](http://www.lintcode.com/en/problem/topological-sorting/)
- [Topological Sorting - GeeksforGeeks](http://www.geeksforgeeks.org/topological-sorting/)
~~~
Given an directed graph, a topological order of the graph nodes is defined as follow:
For each directed edge A -> B in graph, A must before B in the order list.
The first node in the order can be any node in the graph with no nodes direct to it.
Find any topological order for the given graph.
~~~
ExampleFor graph as follow:
~~~
The topological order can be:
[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
...
Note
You can assume that there is at least one topological order in the graph.
Challenge
Can you do it in both BFS and DFS?
~~~
### 题解1 - [DFS](# "Depth-First Search, 深度优先搜索") and [BFS](# "Breadth-First Search, 广度优先搜索")
图搜索相关的问题较为常见的解法是用 [DFS](# "Depth-First Search, 深度优先搜索")，这里结合 [BFS](# "Breadth-First Search, 广度优先搜索") 进行求解，分为三步走：
1. 统计各定点的入度——只需统计节点在邻接列表中出现的次数即可知。
1. 遍历图中各节点，找到入度为0的节点。
1. 对入度为0的节点进行递归 [DFS](# "Depth-First Search, 深度优先搜索")，将节点加入到最终返回结果中。
### C++
~~~
/**
* Definition for Directed graph.
* struct DirectedGraphNode {
* int label;
* vector<DirectedGraphNode *> neighbors;
* DirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
/**
* @param graph: A list of Directed graph node
* @return: Any topological order for the given graph.
*/
vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {
vector<DirectedGraphNode*> result;
if (graph.size() == 0) return result;
map<DirectedGraphNode*, int> indegree;
// get indegree of all DirectedGraphNode
indeg(graph, indegree);
// dfs recursively
for (int i = 0; i < graph.size(); ++i) {
if (indegree[graph[i]] == 0) {
dfs(indegree, graph[i], result);
}
}
return result;
}
private:
/** get indegree of all DirectedGraphNode
*
*/
void indeg(vector<DirectedGraphNode*> &graph,
map<DirectedGraphNode*, int> &indegree) {
for (int i = 0; i < graph.size(); ++i) {
for (int j = 0; j < graph[i]->neighbors.size(); j++) {
if (indegree.find(graph[i]->neighbors[j]) == indegree.end()) {
indegree[graph[i]->neighbors[j]] = 1;
} else {
indegree[graph[i]->neighbors[j]] += 1;
}
}
}
}
void dfs(map<DirectedGraphNode*, int> &indegree, DirectedGraphNode *i,
vector<DirectedGraphNode*> &ret) {
ret.push_back(i);
indegree[i]--;
for (int j = 0; j < i->neighbors.size(); ++j) {
indegree[i->neighbors[j]]--;
if (indegree[i->neighbors[j]] == 0) {
dfs(indegree, i->neighbors[j], ret);
}
}
}
};
~~~
### 源码分析
C++中使用 unordered_map 可获得更高的性能，私有方法中使用引用传值。
### 复杂度分析
以 V 表示顶点数，E 表示有向图中边的条数。
首先获得节点的入度数，时间复杂度为 O(V+E)O(V+E)O(V+E), 使用了哈希表存储，空间复杂度为 O(V)O(V)O(V). 遍历图求得入度为0的节点，时间复杂度为 O(V)O(V)O(V). 仅在入度为0时调用 [DFS](# "Depth-First Search, 深度优先搜索")，故时间复杂度为 O(V+E)O(V+E)O(V+E).
需要注意的是这里的 [DFS](# "Depth-First Search, 深度优先搜索") 不是纯 [DFS](# "Depth-First Search, 深度优先搜索")，使用了 [BFS](# "Breadth-First Search, 广度优先搜索") 的思想进行了优化，否则一个节点将被遍历多次，时间复杂度可能恶化为指数级别。
综上，时间复杂度近似为 O(V+E)O(V+E)O(V+E), 空间复杂度为 O(V)O(V)O(V).
### 题解2 - [BFS](# "Breadth-First Search, 广度优先搜索")
拓扑排序除了可用 [DFS](# "Depth-First Search, 深度优先搜索") 求解外，也可使用 [BFS](# "Breadth-First Search, 广度优先搜索"), 具体方法为：
1. 获得图中各节点的入度。
1. [BFS](# "Breadth-First Search, 广度优先搜索") 首先遍历求得入度数为0的节点，入队，便于下一次 [BFS](# "Breadth-First Search, 广度优先搜索")。
1. 队列不为空时，弹出队顶元素并对其邻接节点进行 [BFS](# "Breadth-First Search, 广度优先搜索")，将入度为0的节点加入到最终结果和队列中，重复此过程直至队列为空。
### C++
~~~
/**
* Definition for Directed graph.
* struct DirectedGraphNode {
* int label;
* vector<DirectedGraphNode *> neighbors;
* DirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
/**
* @param graph: A list of Directed graph node
* @return: Any topological order for the given graph.
*/
vector<DirectedGraphNode*> topSort(vector<DirectedGraphNode*> graph) {
vector<DirectedGraphNode*> result;
if (graph.size() == 0) return result;
map<DirectedGraphNode*, int> indegree;
// get indegree of all DirectedGraphNode
indeg(graph, indegree);
queue<DirectedGraphNode*> q;
// bfs
bfs(graph, indegree, q, result);
return result;
}
private:
/** get indegree of all DirectedGraphNode
*
*/
void indeg(vector<DirectedGraphNode*> &graph,
map<DirectedGraphNode*, int> &indegree) {
for (int i = 0; i < graph.size(); ++i) {
for (int j = 0; j < graph[i]->neighbors.size(); j++) {
if (indegree.find(graph[i]->neighbors[j]) == indegree.end()) {
indegree[graph[i]->neighbors[j]] = 1;
} else {
indegree[graph[i]->neighbors[j]] += 1;
}
}
}
}
void bfs(vector<DirectedGraphNode*> &graph, map<DirectedGraphNode*, int> &indegree,
queue<DirectedGraphNode *> &q, vector<DirectedGraphNode*> &ret) {
for (int i = 0; i < graph.size(); ++i) {
if (indegree[graph[i]] == 0) {
ret.push_back(graph[i]);
q.push(graph[i]);
}
}
while (!q.empty()) {
DirectedGraphNode *cur = q.front();
q.pop();
for(int j = 0; j < cur->neighbors.size(); ++j) {
indegree[cur->neighbors[j]]--;
if (indegree[cur->neighbors[j]] == 0) {
ret.push_back(cur->neighbors[j]);
q.push(cur->neighbors[j]);
}
}
}
}
};
~~~
### 源码分析
C++中在判断入度是否为0时将对 map 产生副作用，在求入度数时只有入度数大于等于1才会出现在 map 中，故不在 map 中时直接调用 indegree 方法将产生新的键值对，初始值为0，恰好满足此题需求。
### 复杂度分析
同题解1 的分析，时间复杂度为 O(V+E)O(V+E)O(V+E), 空间复杂度为 O(V)O(V)O(V).
### Reference
- [Topological Sorting 参考程序 Java/C++/Python](http://www.jiuzhang.com/solutions/topological-sorting/)
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume