给定一个大小为m * n的网格，让我们假设您从(1，1)开始，而您的目标是达到(m，n)。无论如何，如果您在(x，y)上，则可以转到(x，y + 1)或(x + 1，y)。
现在考虑是否在网格中添加了一些障碍。会有多少条独特的路径?
网格中的障碍物和空白区域分别标记为1和0。
例子：
Input: [[0, 0, 0],
[0, 1, 0],
[0, 0, 0]]
Output : 2
There is only one obstacle in the middle.我们讨论了一个问题，当网格中没有障碍物时，该问题用于计算网格中唯一路径的数量。但是这里的情况却大不相同。在网格中移动时，我们会遇到一些无法跳跃的障碍，因此无法到达右下角。
使用动态编程可以解决该问题，这是最有效的解决方案。像每个动态问题概念一样，我们不会重新计算子问题。将构建一个临时2D矩阵，并使用自下而上的方法存储值。
方法
- 创建与给定矩阵大小相同的2D矩阵以存储结果。
- 遍历创建的数组行并开始填充其中的值。
- 如果发现障碍物，则将该值设置为0。
- 对于第一行和第一列，如果未发现障碍物，则将该值设置为1。
- 如果给定母体的相应位置上没有障碍物，则设置右值和上值的总和
- 返回创建的二维矩阵的最后一个值
C++
// C++ code to find number of unique paths
// in a Matrix
#include
using namespace std;
int uniquePathsWithObstacles(vector>& A)
{
int r = A.size(), c = A[0].size();
// create a 2D-matrix and initializing
// with value 0
vector> paths(r, vector(c, 0));
// Initializing the left corner if
// no obstacle there
if (A[0][0] == 0)
paths[0][0] = 1;
// Initializing first column of
// the 2D matrix
for(int i = 1; i < r; i++)
{
// If not obstacle
if (A[i][0] == 0)
paths[i][0] = paths[i-1][0];
}
// Initializing first row of the 2D matrix
for(int j = 1; j < c; j++)
{
// If not obstacle
if (A[0][j] == 0)
paths[0][j] = paths[0][j - 1];
}
for(int i = 1; i < r; i++)
{
for(int j = 1; j < c; j++)
{
// If current cell is not obstacle
if (A[i][j] == 0)
paths[i][j] = paths[i - 1][j] +
paths[i][j - 1];
}
}
// Returning the corner value
// of the matrix
return paths[r - 1];
}
// Driver code
int main()
{
vector> A = { { 0, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 0 } };
cout << uniquePathsWithObstacles(A) << " \n";
}
// This code is contributed by ajaykr00kj Java
// Java code to find number of unique paths
// in a Matrix
public class Main
{
static int uniquePathsWithObstacles(int[][] A)
{
int r = 3, c = 3;
// create a 2D-matrix and initializing
// with value 0
int[][] paths = new int[r];
for(int i = 0; i < r; i++)
{
for(int j = 0; j < c; j++)
{
paths[i][j] = 0;
}
}
// Initializing the left corner if
// no obstacle there
if (A[0][0] == 0)
paths[0][0] = 1;
// Initializing first column of
// the 2D matrix
for(int i = 1; i < r; i++)
{
// If not obstacle
if (A[i][0] == 0)
paths[i][0] = paths[i - 1][0];
}
// Initializing first row of the 2D matrix
for(int j = 1; j < c; j++)
{
// If not obstacle
if (A[0][j] == 0)
paths[0][j] = paths[0][j - 1];
}
for(int i = 1; i < r; i++)
{
for(int j = 1; j < c; j++)
{
// If current cell is not obstacle
if (A[i][j] == 0)
paths[i][j] = paths[i - 1][j] +
paths[i][j - 1];
}
}
// Returning the corner value
// of the matrix
return paths[r - 1];
}
// Driver code
public static void main(String[] args) {
int[][] A = { { 0, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 0 } };
System.out.print(uniquePathsWithObstacles(A));
}
}
// This code is contributed by divyeshrabadiya07.Python
# Python code to find number of unique paths in a
# matrix with obstacles.
def uniquePathsWithObstacles(A):
# create a 2D-matrix and initializing with value 0
paths = [[0]*len(A[0]) for i in A]
# initializing the left corner if no obstacle there
if A[0][0] == 0:
paths[0][0] = 1
# initializing first column of the 2D matrix
for i in range(1, len(A)):
# If not obstacle
if A[i][0] == 0:
paths[i][0] = paths[i-1][0]
# initializing first row of the 2D matrix
for j in range(1, len(A[0])):
# If not obstacle
if A[0][j] == 0:
paths[0][j] = paths[0][j-1]
for i in range(1, len(A)):
for j in range(1, len(A[0])):
# If current cell is not obstacle
if A[i][j] == 0:
paths[i][j] = paths[i-1][j] + paths[i][j-1]
# returning the corner value of the matrix
return paths[-1][-1]
# Driver Code
A = [[0, 0, 0], [0, 1, 0], [0, 0, 0]]
print(uniquePathsWithObstacles(A))C#
// C# code to find number of unique paths
// in a Matrix
using System;
class GFG {
static int uniquePathsWithObstacles(int[,] A)
{
int r = 3, c = 3;
// create a 2D-matrix and initializing
// with value 0
int[,] paths = new int[r,c];
for(int i = 0; i < r; i++)
{
for(int j = 0; j < c; j++)
{
paths[i, j] = 0;
}
}
// Initializing the left corner if
// no obstacle there
if (A[0, 0] == 0)
paths[0, 0] = 1;
// Initializing first column of
// the 2D matrix
for(int i = 1; i < r; i++)
{
// If not obstacle
if (A[i, 0] == 0)
paths[i, 0] = paths[i - 1, 0];
}
// Initializing first row of the 2D matrix
for(int j = 1; j < c; j++)
{
// If not obstacle
if (A[0, j] == 0)
paths[0, j] = paths[0, j - 1];
}
for(int i = 1; i < r; i++)
{
for(int j = 1; j < c; j++)
{
// If current cell is not obstacle
if (A[i, j] == 0)
paths[i, j] = paths[i - 1, j] +
paths[i, j - 1];
}
}
// Returning the corner value
// of the matrix
return paths[r - 1, c - 1];
}
// Driver code
static void Main() {
int[,] A = { { 0, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 0 } };
Console.WriteLine(uniquePathsWithObstacles(A));
}
}
// This code is contributed by divyesh072019.Javascript
输出
2