## [LeetCode] Unique Paths II

### Unique Paths IIFollow 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.

```class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
if(m<=0){
return 0;
}
int n = obstacleGrid[0].size();
if(n<=0){
return 0;
}
if(obstacleGrid[0][0]==1 || obstacleGrid[m-1][n-1]==1){
return 0;
}

int result = 0;

helper(obstacleGrid, 0, 0, m, n, result);

return result;
}

void helper(vector<vector<int>>& obstacleGrid, int i, int j, int m, int n, int& result){
if(obstacleGrid[i][j]==1){
return;
}
if(i==m-1&&j==n-1){
result++;
return;
}
if(i<m-1){
helper(obstacleGrid, i+1, j, m, n, result);
}
if(j<n-1){
helper(obstacleGrid, i, j+1, m, n, result);
}
}
};```

d[i][j]=0, 当matrix[i][j]=1;

d[i][j]=d[i-1][j] + d[i][j-1]; 其他。

```class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
if(m<=0){
return 0;
}
int n = obstacleGrid[0].size();
if(n<=0){
return 0;
}
if(obstacleGrid[0][0]==1 || obstacleGrid[m-1][n-1]==1){
return 0;
}
vector<vector<int>> d(m, vector<int>(n, 0));
d[0][0] = 1;
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(obstacleGrid[i][j]==1){
d[i][j]=0;
}else{
if(i>0){
d[i][j] += d[i-1][j];
}
if(j>0){
d[i][j] += d[i][j-1];
}
}
}
}
return d[m-1][n-1];
}
};```