63. Unique Paths 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.

相比62. Unique Paths 就是增加了障碍物，递归公式为：

if(obstacleGrid[i][j]==1)

dp[i][j]=0;

else

dp[i][j]=dp[i][j-1]+dp[i-1][j];

但是要注意下边界条件的处理。

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