我在一次采访中被问到这个问题，并且一直在努力寻找一个优雅的解决方案（用 C 语言），问题陈述：
在 C 'numberOfPaths' 中写一个函数，它接受上述二维数组，返回从左上角单元格到右下角单元格的有效路径数（即 [0,0] 到 [M-1 ,N-1])。
编辑：忘了提到要求是递归解决方案
帮助将不胜感激！ 谢谢
【问题讨论】：
如果您正在寻找递归解决方案，您可以使用 DFS。
DFS (array, x, y)
{
if (array [x][y]==0 || x>M || y>N){
return;
}
if (x==M && y==N){
count++;
return;
}
DFS (array, x, y+1);
DFS (array, x+1, y);
}
【讨论】：
到给定点的路径数就是到上面点的路径数加上到左边点的路径数。所以，伪代码大概是：
num_paths[0][0] = 1;
for (x = 0; x < M; ++x)
for (y = 0; y < N; ++y)
if (!allowed_through[x][y])
num_paths[x][y] = 0;
else
num_paths[x][y] = num_paths[x-1][y] + num_paths[x][y-1];
您需要 x=0 和 y=0 的特殊情况，否则，我认为应该这样做。
【讨论】：
#include <stdio.h>
int count=0;
int maxrows = 10;
int maxcols = 10;
int M, N;
void DFS (int array[][10], int x, int y)
{
int r, c;
/* process element at input row and column */
if (array [x][y]==0 || x>M || y>N){
/* no path forward; return */
return;
}
if (x==M-1 && y==N-1){
/* found path; increment count */
count++;
return;
}
/* recurse: to matrix starting from same row, next column */
r = x;
c = y +1;
if (c < N-1) {
DFS (array, r,c);
} else {
/* if last column - check to see */
/* if rest of rows in last column allow for a path */
int tr = r;
while ( tr <= M-1) {
if (array[tr][c] == 1) {
tr++;
}
else {
return;
}
}
/* reached last node - path exists! */
count++;
}
/* recurse: to matrix starting from next row, same column */
r = x+1;
c = y;
if (r < M-1) {
DFS (array, r,c);
} else {
/* if last row - check to see */
/* if rest of columns in last row allow for a path */
int tc = c;
while ( tc <= N-1) {
if (array[r][tc] == 1) {
tc++;
} else {
return;
}
}
/* reached last node - path exists! */
count++;
}
}
int main () {
int i, j;
scanf("%d %d",&M,&N);
int a[10][10] = {};
int row, col;
for(i=0;i<M;i++)
for(j=0;j<N;j++)
scanf("%d", &a[i][j]);
if ((M > maxrows) || (N > maxcols)) {
printf("max of 10 rows and 10 cols allowed for input\n");
return (-1);
};
/* print input matrix */
for(row=0;row<M;row++) {
for(col=0;col<N;col++){
printf("%d ",a[row][col]);
}
printf(" EOR\n");
}
DFS(a,0,0);
printf("number of paths is %d\n", count);
return 0;
}
【讨论】：
在打印所有路径之前，请先尝试此功能。 如果向量 Out 的大小为 0 则路径数为 0，但如果 size(Out) > 0 则向量节点的大小 + 1 为从左上角到右下角的路径总数。
#include <iostream>
#include <vector>
using namespace std;
typedef vector<pair<int,int> > vPii;
bool pathTL2BR( int Arr2D[][4], vPii &Out, vPii &Nodes,
int _x,int _y, int _M, int _N)
{
bool out1 = false;
bool out2 = false;
if( Arr2D[_x][_y] == 1 )
{
if( _y+1 < _N )
out1 = pathTL2BR( Arr2D, Out, Nodes, _x, _y+1, _M, _N);
if( _x+1 < _M )
out2 = pathTL2BR( Arr2D, Out, Nodes, _x+1, _y, _M, _N);
if( (out1 || out2) ||
( (_x == (_M-1)) && (_y == (_N-1)) ) )
{
if(out1 && out2)
Nodes.push_back( make_pair(_x,_y ) );
Out.push_back( make_pair(_x,_y ) );
return true;
}
else
return false;
}
else
return false;
}
// Driver program to test above function
int main()
{
int Arr2D[][4] = {
{1,1,1,1},
{0,1,0,1},
{0,1,0,1},
{0,1,0,1}
};
vPii Out;
vPii Nodes;
vector<vPii> Output;
pathTL2BR( Arr2D, Out, Nodes, 0, 0, 4, 4);
return 0;
}
【讨论】：
这是一个 python 解决方案，我已经在 cmets 中做了解释。
def find_num_paths(arr_2D, i, j):
# i,j is the start point and you have to travel all the way back to 0,0
if i == j and i == 0:
return 1 # you have reached the start point
if i < 0 or j < 0 or arr_2D[i][j] == 0: # out of range or no path from that point
return 0
if arr_2D[i][j] == 1:
return find_num_paths(arr_2D, i, j-1) + find_num_paths(arr_2D, i-1, j) + find_num_paths(arr_2D, i-1, j-1) # you could go one step above, to the left or diagonally up.
【讨论】：