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Now consider if some obstacles are added to the grids. How many unique paths would there be?
Example 1:
Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right
class
Solution_63_1 {
public:
int
uniquePathsWithObstacles(vector
<vector
<
int
>>& obstacleGrid) {
const
int m
= obstacleGrid.
size();
const
int n
= obstacleGrid[
0].
size();
if (obstacleGrid[
0][
0]
|| obstacleGrid[m
-
1][n
-
1])
return
0;
vector
<
int
>
f(n,
0);
f[
0]
=
1;
for (
int i
=
0; i
!= m;
++i) {
//先判断每一行的第一个
//if (obstacleGrid[i][0] == 1)
// f[0] = 0;
//else
// f[0] = f[0];
if (f[
0]
==
0)
f[
0]
=
0;
else
if (obstacleGrid[i][
0]
==
1)
f[
0]
=
0;
else
f[
0]
=
1;
for (
int j
=
1; j
!= n;
++j) {
if (obstacleGrid[i][j]
==
1)
f[j]
=
0;
else
f[j]
= f[j]
+ f[j
-
1];
}
}
return f[n
-
1];
}
};
//优化一下代码
class
Solution_63_2 {
public:
int
uniquePathsWithObstacles(vector
<vector
<
int
>>& obstacleGrid) {
const
int m
= obstacleGrid.
size();
const
int n
= obstacleGrid[
0].
size();
if (obstacleGrid[
0][
0]
|| obstacleGrid[m
-
1][n
-
1])
return
0;
vector
<
int
>
f(n,
0);
f[
0]
=
1;
for (
int i
=
0; i
!= m;
++i) {
f[
0]
= f[
0]
==
0
?
0
: (obstacleGrid[i][
0]
?
0
:
1);
for (
int j
=
1; j
!= n;
++j) {
f[j]
= obstacleGrid[i][j]
?
0
: f[j]
+ f[j
-
1];
}
}
return f[n
-
1];
}
};
class
Solution_63_3 {
public:
int
uniquePathsWithObstacles(vector
<vector
<
int
>>& obstacleGrid) {
const
int m
= obstacleGrid.
size();
if (m
<=
0)
return
0;
const
int n
= obstacleGrid[
0].
size();
if (n
<=
0)
return
0;
if (obstacleGrid[
0][
0]
|| obstacleGrid[m
-
1][n
-
1])
return
0;
int f[
101][
101];
f[
0][
0]
=
1;
//初始化第一行
for (
int i
=
1; i
!= n;
++i) {
if (obstacleGrid[
0][i]
==
1) {
f[
0][i]
=
0;
}
else {
f[
0][i]
= f[
0][i
-
1];
}
}
//初始化第一列
for (
int j
=
1; j
!= m;
++j) {
if (obstacleGrid[j][
0]
==
1) {
f[j][
0]
=
0;
}
else {
f[j][
0]
= f[j
-
1][
0];
}
}
for (
int i
=
1; i
!= m;
++i) {
for (
int j
=
1; j
!= n;
++j) {
if (obstacleGrid[i][j]
==
1)
f[i][j]
=
0;
else
f[i][j]
= f[i
-
1][j]
+ f[i][j
-
1];
}
}
return f[m
-
1][n
-
1];
}
};
//深搜 缓存
class
Solution_63_4 {
public:
int
uniquePathsWithObstacles(
const vector
<vector
<
int
>
>& obstacleGrid) {
const
int m
= obstacleGrid.
size();
const
int n
= obstacleGrid[
0].
size();
if (obstacleGrid[
0][
0]
|| obstacleGrid[m
-
1][n
-
1])
return
0;
f
= vector
<vector
<
int
>
>(m, vector
<
int
>(n,
0));
f[
0][
0]
=
1;
return
dfs(obstacleGrid, m
-
1, n
-
1);
}
private:
vector
<vector
<
int
>
> f;
// 缓存
// @return 从(0, 0) 到(x, y)的路径总数
int
dfs(
const vector
<vector
<
int
>
>& obstacleGrid,
int x,
int y) {
if (x
<
0
|| y
<
0)
return
0;
//数据非法，终止条件
// (x,y) 是障碍
if (obstacleGrid[x][y])
return
0;
if (x
==
0
and y
==
0)
return f[
0][
0];
//回到起点，收敛条件
if (f[x][y]
>
0) {
return f[x][y];
}
else {
return f[x][y]
=
dfs(obstacleGrid, x
-
1, y)
+
dfs(obstacleGrid, x, y
-
1);
}
}
};