63. Unique Paths II

Description：

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.

Solution：

这是Unique Path的进化版，在路途中设置了障碍，即某些位置不能走。

标志1的位置不可以走，标号0的位置可以走，从位置[0][0]到位置[m-1][n-1]。

解决方法同理，在第一行、第一列的所有位置标记路径数为1，有障碍的位置，标记路径数为0，剩下的每个位置的路径数为它的左边和上边位置路径数之和。

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