LeetCode | Unique Paths II（唯一路径II）

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 3x3 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.

题目解析：

这道题是上一题的扩展：LeetCode | Unique Paths

注意的是先初始化第一行和第一列，从左到右初始化的过程中，碰到一个结点有障碍物，那么从该点开始之后的路径值都为0。

当遍历内部结点的时候，没有障碍物就是左边和上面两个路径值的和。不用刻意判断左边和上面是否有障碍物，有的话路径值已经赋为0了，直接相加即可。

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
vector<vector<int> >res;
int row = obstacleGrid.size();
if(row == 0)
return 0;
int col = obstacleGrid[0].size();
if(col == 0)
return 0;
for(int i = 0;i < row;i++){
vector<int> tmp = vector<int> (col,0);
res.push_back(tmp);
}
//一开始就有阻碍物的话，就一定不可达
if(obstacleGrid[0][0] == 1)
return 0;
//初始化，第一行和第一列，当碰到有一个障碍物的时候，后面的或下面的就全为0
for(int i = 0;i < row;i++){
if(obstacleGrid[i][0] == 0){
res[i][0] = 1;
}else
break;
}
for(int j = 0;j < col;j++){
if(obstacleGrid[0][j] == 0)
res[0][j] = 1;
else
break;
}
//如果这个位置没有障碍物，那么其路径为右边过来的和上面过来的之和，
//如果有障碍物就为0，因为已经初始化了，就不用赋值了
for(int i = 1;i < row;i++){
for(int j = 1;j < col;j++){
if(obstacleGrid[i][j] == 0)
res[i][j] = res[i-1][j] + res[i][j-1];
}
}
return res[row-1][col-1];
}
};