算法题（查找矩阵中两个单元之间是否存在路径）

本文概述

Java
Python3
C#
C ++
Java
Python3

给定N X N矩阵, 其中用1, 0, 2, 3填充。查找是否存在从源到目标的路径, 仅遍历空白单元格。你可以上下左右移动。

单元格的值1表示来源。
单元格的值2表示目的地。
单元格的值3表示空白单元格。
单元格的值0表示空白墙。

注意：只有一个来源和一个目的地(接收器)。
例子：

输入：M [3] [3] = {{0, 3, 2}, {3, 3, 0}, {1, 3, 0}};
输出：是
说明：输入：M [4] [4] = {{0, 3, 1, 0}, {3, 0, 3, 3}, {2, 3, 0, 3}, {0, 3 , 3, 3}};
输出：是
说明：

简单的解决方案：递归。
方法：
在每个矩阵中找到单元的源索引, 然后递归地找到从源索引到矩阵中目标的路径。该算法涉及递归查找所有路径, 直到找到到达目的地的最终路径。
算法：

遍历矩阵并找到矩阵的起始索引。
创建一个采用索引和访问矩阵的递归函数。
标记当前单元格, 然后检查当前单元格是否为目的地。如果当前单元格是目的地, 则返回true。
为所有相邻的空白和未访问单元格调用递归函数。
如果任何递归函数返回true, 则取消标记该单元格并返回true, 否则取消标记该单元格并返回false。

Java

//Java program to find path between two //cell in matrix class Path { //Method for finding and printing //whether the path exists or not public static void isPath( int matrix[][], int n) { //Defining visited array to keep //track of already visited indexes boolean visited[][] = new boolean [n][n]; //Flag to indicate whether the //path exists or not boolean flag = false ; for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { //if matrix[i][j] is source //and it is not visited if ( matrix[i][j] == 1 & & !visited[i][j]) //Starting from i, j and //then finding the path if (isPath( matrix, i, j, visited)) { //if path exists flag = true ; break ; } } } if (flag) System.out.println( "YES" ); else System.out.println( "NO" ); } //Method for checking boundries public static boolean isSafe( int i, int j, int matrix[][]) { if ( i> = 0 & & i < matrix.length & & j> = 0 & & j < matrix[ 0 ].length) return true ; return false ; } //Returns true if there is a //path from a source (a //cell with value 1) to a //destination (a cell with //value 2) public static boolean isPath( int matrix[][], int i, int j, boolean visited[][]) { //Checking the boundries, walls and //whether the cell is unvisited if ( isSafe(i, j, matrix) & & matrix[i][j] != 0 & & !visited[i][j]) { //Make the cell visited visited[i][j] = true ; //if the cell is the required //destination then return true if (matrix[i][j] == 2 ) return true ; //traverse up boolean up = isPath( matrix, i - 1 , j, visited); //if path is found in up //direction return true if (up) return true ; //traverse left boolean left = isPath( matrix, i, j - 1 , visited); //if path is found in left //direction return true if (left) return true ; //traverse down boolean down = isPath( matrix, i + 1 , j, visited); //if path is found in down //direction return true if (down) return true ; //traverse right boolean right = isPath( matrix, i, j + 1 , visited); //if path is found in right //direction return true if (right) return true ; } //no path has been found return false ; } //driver program to //check above function public static void main(String[] args) { int matrix[][] = { { 0 , 3 , 0 , 1 }, { 3 , 0 , 3 , 3 }, { 2 , 3 , 3 , 3 }, { 0 , 3 , 3 , 3 } }; //calling isPath method isPath(matrix, 4 ); } } /* This code is contributed by Madhu Priya */

Python3

# Python3 program to find # path between two cell in matrix # Method for finding and printing # whether the path exists or not def isPath(matrix, n): # Defining visited array to keep # track of already visited indexes visited = [[ False for x in range (n)] for y in range (n)]# Flag to indicate whether the # path exists or not flag = False for i in range (n): for j in range (n):# If matrix[i][j] is source # and it is not visited if (matrix[i][j] = = 1 and not visited[i][j]): # Starting from i, j and # then finding the path if (checkPath(matrix, i, j, visited)):# If path exists flag = True break if (flag): print ( "YES" ) else : print ( "NO" ) # Method for checking boundries def isSafe(i, j, matrix):if (i> = 0 and i < len (matrix) and j> = 0 and j < len (matrix[ 0 ])): return True return False # Returns true if there is a # path from a source(a # cell with value 1) to a # destination(a cell with # value 2) def checkPath(matrix, i, j, visited): # Checking the boundries, walls and # whether the cell is unvisited if (isSafe(i, j, matrix) and matrix[i][j] ! = 0 and not visited[i][j]):# Make the cell visited visited[i][j] = True # If the cell is the required # destination then return true if (matrix[i][j] = = 2 ): return True # traverse up up = checkPath(matrix, i - 1 , j, visited) # If path is found in up # direction return true if (up): return True # Traverse left left = checkPath(matrix, i, j - 1 , visited) # If path is found in left # direction return true if (left): return True # Traverse down down = checkPath(matrix, i + 1 , j, visited) # If path is found in down # direction return true if (down): return True # Traverse right right = checkPath(matrix, i, j + 1 , visited) # If path is found in right # direction return true if (right): return True# No path has been found return False # Driver code if __name__ = = "__main__" :matrix = [[ 0 , 3 , 0 , 1 ], [ 3 , 0 , 3 , 3 ], [ 2 , 3 , 3 , 3 ], [ 0 , 3 , 3 , 3 ]] # calling isPath method isPath(matrix, 4 ) # This code is contributed by Chitranayal

//C# program to find path between two //cell in matrix using System; class GFG{ //Method for finding and printing //whether the path exists or not static void isPath( int [, ] matrix, int n) {//Defining visited array to keep //track of already visited indexes bool [, ] visited = new bool [n, n]; //Flag to indicate whether the //path exists or not bool flag = false ; for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) {//If matrix[i][j] is source //and it is not visited if (matrix[i, j] == 1 & & !visited[i, j])//Starting from i, j and //then finding the path if (isPath(matrix, i, j, visited)) {//If path exists flag = true ; break ; } } } if (flag) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } //Method for checking boundries public static bool isSafe( int i, int j, int [, ] matrix) { if (i> = 0 & & i < matrix.GetLength(0) & & j> = 0 & & j < matrix.GetLength(1)) return true ; return false ; } //Returns true if there is a path from //a source (a cell with value 1) to a //destination (a cell with value 2) public static bool isPath( int [, ] matrix, int i, int j, bool [, ] visited) {//Checking the boundries, walls and //whether the cell is unvisited if (isSafe(i, j, matrix) & & matrix[i, j] != 0 & & !visited[i, j]) {//Make the cell visited visited[i, j] = true ; //If the cell is the required //destination then return true if (matrix[i, j] == 2) return true ; //Traverse up bool up = isPath(matrix, i - 1, j, visited); //If path is found in up //direction return true if (up) return true ; //Traverse left bool left = isPath(matrix, i, j - 1, visited); //If path is found in left //direction return true if (left) return true ; //Traverse down bool down = isPath(matrix, i + 1, j, visited); //If path is found in down //direction return true if (down) return true ; //Traverse right bool right = isPath(matrix, i, j + 1, visited); //If path is found in right //direction return true if (right) return true ; }//No path has been found return false ; } //Driver code static void Main() { int [, ] matrix = { { 0, 3, 0, 1 }, { 3, 0, 3, 3 }, { 2, 3, 3, 3 }, { 0, 3, 3, 3 } }; //Calling isPath method isPath(matrix, 4); } } //This code is contributed by divyeshrabadiya07

输出如下：

YES

复杂度分析：

时间复杂度：O(4n * m)。
对于每个单元, 可以有4个相邻的未访问单元, 因此时间复杂度为O(4n * m)。
空间复杂度：O(n * m)
需要空间来存储访问数组。

高效的解决方案：
图形。
方法：
这个想法是使用
广度优先搜索
。将每个单元视为一个节点, 并将任何两个相邻单元之间的每个边界作为一条边。所以Node的总数是N *N。
【算法题（查找矩阵中两个单元之间是否存在路径）】因此, 想法是从起始单元格开始进行广度优先搜索, 直到找到终止单元格。
算法：

创建一个具有N * N个节点(顶点)的空图, 将所有节点推入图中, 并记下源顶点和宿顶点。
现在在图上应用BFS, 创建一个队列并将源节点插入该队列中
运行循环, 直到队列大小大于0
删除队列的最前面的节点, 如果目标返回true, 则检查该节点是否为目标。标记节点
检查所有相邻的单元格(如果未访问), 并将它们空白插入队列。
如果未到达目的地, 则返回true。

C ++

//C++ program to find path //between two cell in matrix #include < bits/stdc++.h> using namespace std; #define N 4 class Graph { int V; list< int> * adj; public : Graph( int V) { this -> V = V; adj = new list< int> [V]; } void addEdge( int s, int d); bool BFS( int s, int d); }; //add edge to graph void Graph::addEdge( int s, int d) { adj展开.push_back(d); } //BFS function to find path //from source to sink bool Graph::BFS( int s, int d) { //Base case if (s == d) return true ; //Mark all the vertices as not visited bool * visited = new bool [V]; for ( int i = 0; i < V; i++) visited[i] = false ; //Create a queue for BFS list< int> queue; //Mark the current node as visited and //enqueue it visited展开 = true ; queue.push_back(s); //it will be used to get all adjacent //vertices of a vertex list< int> ::iterator i; while (!queue.empty()) { //Dequeue a vertex from queue s = queue.front(); queue.pop_front(); //Get all adjacent vertices of the //dequeued vertex s. If a adjacent has //not been visited, then mark it visited //and enqueue it for ( i = adj展开.begin(); i != adj展开.end(); ++i) { //If this adjacent node is the //destination node, then return true if (*i == d) return true ; //Else, continue to do BFS if (!visited[*i]) { visited[*i] = true ; queue.push_back(*i); } } } //If BFS is complete without visiting d return false ; } bool isSafe( int i, int j, int M[][N]) { if ( (i < 0 || i> = N) || (j < 0 || j> = N) || M[i][j] == 0) return false ; return true ; } //Returns true if there is //a path from a source (a //cell with value 1) to a //destination (a cell with //value 2) bool findPath( int M[][N]) { //source and destination int s, d; int V = N * N + 2; Graph g(V); //create graph with n*n node //each cell consider as node //Number of current vertex int k = 1; for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { if (M[i][j] != 0) { //connect all 4 adjacent //cell to current cell if (isSafe(i, j + 1, M)) g.addEdge(k, k + 1); if (isSafe(i, j - 1, M)) g.addEdge(k, k - 1); if (i < N - 1 & & isSafe(i + 1, j, M)) g.addEdge(k, k + N); if (i> 0 & & isSafe(i - 1, j, M)) g.addEdge(k, k - N); } //Source index if (M[i][j] == 1) s = k; //Destination index if (M[i][j] == 2) d = k; k++; } } //find path Using BFS return g.BFS(s, d); } //driver program to check //above function int main() { int M[N][N] = { { 0, 3, 0, 1 }, { 3, 0, 3, 3 }, { 2, 3, 3, 3 }, { 0, 3, 3, 3 } }; (findPath(M) == true ) ? cout < < "Yes" : cout < < "No" < < endl; return 0; }

Java

//Java program to find path between two //cell in matrix import java.util.*; class Graph { int V; List< List< Integer> > adj; Graph( int V) { this .V = V; adj = new ArrayList< > (V); for ( int i = 0 ; i < V; i++) { adj.add(i, new ArrayList< > ()); } } //add edge to graph void addEdge( int s, int d) { adj.get(s).add(d); } //BFS function to find path //from source to sink boolean BFS( int s, int d) { //Base case if (s == d) return true ; //Mark all the vertices as not visited boolean [] visited = new boolean [V]; //Create a queue for BFS Queue< Integer> queue = new LinkedList< > (); //Mark the current node as visited and //enqueue it visited展开 = true ; queue.offer(s); //it will be used to get all adjacent //vertices of a vertex List< Integer> edges; while (!queue.isEmpty()) { //Dequeue a vertex from queue s = queue.poll(); //Get all adjacent vertices of the //dequeued vertex s. If a adjacent has //not been visited, then mark it visited //and enqueue it edges = adj.get(s); for ( int curr : edges) { //If this adjacent node is the //destination node, then return true if (curr == d) return true ; //Else, continue to do BFS if (!visited[curr]) { visited[curr] = true ; queue.offer(curr); } } } //If BFS is complete without visiting d return false ; } static boolean isSafe( int i, int j, int [][] M) { int N = M.length; if ( (i < 0 || i> = N) || (j < 0 || j> = N) || M[i][j] == 0 ) return false ; return true ; } //Returns true if there is a //path from a source (a //cell with value 1) to a //destination (a cell with //value 2) static boolean findPath( int [][] M) { //Source and destination int s = - 1 , d = - 1 ; int N = M.length; int V = N * N + 2 ; Graph g = new Graph(V); //Create graph with n*n node //each cell consider as node int k = 1 ; //Number of current vertex for ( int i = 0 ; i < N; i++) { for ( int j = 0 ; j < N; j++) { if (M[i][j] != 0 ) { //connect all 4 adjacent //cell to current cell if (isSafe(i, j + 1 , M)) g.addEdge(k, k + 1 ); if (isSafe(i, j - 1 , M)) g.addEdge(k, k - 1 ); if (i < N - 1 & & isSafe(i + 1 , j, M)) g.addEdge(k, k + N); if (i> 0 & & isSafe(i - 1 , j, M)) g.addEdge(k, k - N); } //source index if (M[i][j] == 1 ) s = k; //destination index if (M[i][j] == 2 ) d = k; k++; } } //find path Using BFS return g.BFS(s, d); } //Driver program to check above function public static void main( String[] args) throws Exception { int [][] M = { { 0 , 3 , 0 , 1 }, { 3 , 0 , 3 , 3 }, { 2 , 3 , 3 , 3 }, { 0 , 3 , 3 , 3 } }; System.out.println( ((findPath(M)) ? "Yes" : "No" )); } } //This code is contributed by abhay379201

Python3

# Python3 program to find path between two # cell in matrix from collections import defaultdict class Graph: def __init__( self ): self .graph = defaultdict( list )# add edge to graph def addEdge( self , u, v): self .graph[u].append(v) # BFS function to find path from source to sink def BFS( self , s, d):# Base case if s = = d: return True# Mark all the vertices as not visited visited = [ False ] * ( len ( self .graph) + 1 ) # Create a queue for BFS queue = [] queue.append(s) # Mark the current node as visited and # enqueue it visited展开 = True while (queue): # Dequeue a vertex from queue s = queue.pop( 0 ) # Get all adjacent vertices of the # dequeued vertex s. If a adjacent has # not been visited, then mark it visited # and enqueue it for i in self .graph展开:# If this adjacent node is the destination # node, then return true if i = = d: return True # Else, continue to do BFS if visited[i] = = False : queue.append(i) visited[i] = True # If BFS is complete without visiting d return False def isSafe(i, j, matrix): if i> = 0 and i < = len (matrix) and j> = 0 and j < = len (matrix[ 0 ]): return True else : return False # Returns true if there is a path from a source (a # cell with value 1) to a destination (a cell with # value 2) def findPath(M): s, d = None , None # source and destination N = len (M) g = Graph() # create graph with n * n node # each cell consider as node k = 1 # Number of current vertex for i in range (N): for j in range (N): if (M[i][j] ! = 0 ): # connect all 4 adjacent cell to # current cell if (isSafe(i, j + 1 , M)): g.addEdge(k, k + 1 ) if (isSafe(i, j - 1 , M)): g.addEdge(k, k - 1 ) if (isSafe(i + 1 , j, M)): g.addEdge(k, k + N) if (isSafe(i - 1 , j, M)): g.addEdge(k, k - N)if (M[i][j] = = 1 ): s = k # destination index if (M[i][j] = = 2 ): d = k k + = 1 # find path Using BFS return g.BFS(s, d) # Driver code if __name__ = = '__main__' : M = [[ 0 , 3 , 0 , 1 ], [ 3 , 0 , 3 , 3 ], [ 2 , 3 , 3 , 3 ], [ 0 , 3 , 3 , 3 ]] if findPath(M): print ( "Yes" ) else : print ( "No" ) # This Code is Contributed by Vikash Kumar 37

输出如下：