原文:
给定一个 mat[][] 、一个源“ s 和一个目的地“ d ”，打印从给定的“ s ”到“ d 的所有唯一路径。从每个单元格中，您可以只向右或向下移动。
示例:
输入: mat[][] = {{1，2，3}，{4，5，6}}，s[] = {0，0}，d[]={1，2} 输出: 1 4 5 6 1 2 5 6 1 2 3 6
输入 : mat[][] = {{1，2}，{3，4}}，s[] = {0，1}，d[] = {1，1} 输出: 2 4
方法:使用从源开始，从矩阵路径中的每个单元格先右后下移动 mat[][] ，并将每个值存储在一个向量中。如果到达目的地，打印矢量作为可能的路径之一。按照以下步骤解决问题:
- 如果当前单元格超出边界，则返回。
- 将当前单元格的值推入向量路径[]。
- 如果当前单元格是目标，则打印当前路径。
- 对值 {i 1，j} 和 {i，j 1}调用相同的函数。
下面是上述方法的实现。
c
// c program for the above approach
#include
using namespace std;
vector > mat;
vector s;
vector d;
int m = 2, n = 3;
// function to print all the paths
void printvector(vector path)
{
int cnt = path.size();
for (int i = 0; i < cnt; i = 2)
cout << mat[path[i]][path[i 1]]
<< " ";
cout << endl;
}
// function to find all the paths recursively
void countpaths(int i, int j, vector path)
{
// base case
if (i > d[0] || j > d[1])
return;
path.push_back(i);
path.push_back(j);
// destination is reached
if (i == d[0] && j == d[1]) {
printvector(path);
return;
}
// calling the function
countpaths(i, j 1, path);
countpaths(i 1, j, path);
}
// drivercode
int main()
{
mat = { { 1, 2, 3 },
{ 4, 5, 6 } };
s = { 0, 0 };
d = { 1, 2 };
vector path;
countpaths(s[0], s[1], path);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.util.*;
class gfg
{
static vector s = new vector<>();
static vector d = new vector<>();
static int m = 2, n = 3;
// function to print all the paths
static void printvector(vector path, int[][] mat)
{
int cnt = path.size();
for (int i = 0; i < cnt; i = 2)
system.out.print(mat[path.get(i)][path.get(i 1)] " ");
system.out.println();
}
// function to find all the paths recursively
static void countpaths(int i, int j, vector path, int[][]mat)
{
// base case
if (i > d.get(0) || j > d.get(1))
return;
path.add(i);
path.add(j);
// destination is reached
if (i == d.get(0) && j == d.get(1)) {
printvector(path,mat);
path.remove(path.size()-1);
path.remove(path.size()-1);
return;
}
// calling the function
countpaths(i, j 1, path,mat);
countpaths(i 1, j, path,mat);
path.remove(path.size()-1);
path.remove(path.size()-1);
}
// drivercode
public static void main(string[] args) {
int[][] mat = {{ 1, 2, 3 },
{ 4, 5, 6 } };
s.add(0);
s.add(0);
d.add(1);
d.add(2);
vector path = new vector<>();
countpaths(s.get(0), s.get(1), path, mat);
}
}
// this code is contributed by rajput-ji.
python 3
# python code for the above approach
mat = none
s = none
d = none
m = 2
n = 3
# function to print all the paths
def printvector(path):
cnt = len(path)
for i in range(0, cnt, 2):
print(mat[path[i]][path[i 1]], end=" ")
print("")
# function to find all the paths recursively
def countpaths(i, j, path):
# base case
if (i > d[0] or j > d[1]):
return
path.append(i)
path.append(j)
# destination is reached
if (i == d[0] and j == d[1]):
printvector(path)
path.pop()
path.pop()
return
# calling the function
countpaths(i, j 1, path)
countpaths(i 1, j, path)
path.pop()
path.pop()
# drivercode
mat = [[1, 2, 3],
[4, 5, 6]]
s = [0, 0]
d = [1, 2]
path = []
countpaths(s[0], s[1], path)
# this code is contributed by saurabh jaiswal
java 描述语言
output:
1 2 3 6
1 2 5 6
1 4 5 6
时间复杂度:o(2n m) 辅助空间: o(1)
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