算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Unique Paths II


### 描述

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a `3 × 3` grid as illustrated below.

```
[
[0,0,0],
[0,1,0],
[0,0,0]
]
```

The total number of unique paths is 2.

Note: `m` and `n` will be at most 100.


### 备忘录法

在上一题的基础上改一下即可。相比动规,简单得多。

{% if book.java %}
{% codesnippet "./code/unique-paths-ii-1."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Unique Paths II
// 深搜 + 缓存,即备忘录法
class Solution {
public:
int uniquePathsWithObstacles(const vector >& 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 >(m, vector(n, 0));
f[0][0] = obstacleGrid[0][0] ? 0 : 1;
return dfs(obstacleGrid, m - 1, n - 1);
}
private:
vector > f; // 缓存

// @return 从 (0, 0) 到 (x, y) 的路径总数
int dfs(const vector >& 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);
}
}
};
```
{% endif %}


### 动规

与上一题类似,但要特别注意第一列的障碍。在上一题中,第一列全部是1,但是在这一题中不同,第一列如果某一行有障碍物,那么后面的行全为0。

{% if book.java %}
{% codesnippet "./code/unique-paths-ii-2."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Unique Paths II
// 动规,滚动数组
// 时间复杂度O(n^2),空间复杂度O(n)
class Solution {
public:
int uniquePathsWithObstacles(vector > &obstacleGrid) {
const int m = obstacleGrid.size();
const int n = obstacleGrid[0].size();
if (obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) return 0;

vector f(n, 0);
f[0] = obstacleGrid[0][0] ? 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];
}
};
```
{% endif %}


### 相关题目

* [Unique Paths](unique-paths.md)
* [Minimum Path Sum](../dp/minimum-path-sum.md)