[Leetcode 58] 63 Unique Path II

Problem:

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.

Analysis:

Simple DP with some more constraints.

If a gird is 1, then there's no way to get to this position, the value should be directly set to 0.

Pay attention to the first inialize of the first row and first column value.

Code:

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        int tab[m][n];
        
        if (obstacleGrid[0][0] == 0)
            tab[0][0] = 1;
        else
            tab[0][0] = 0;
        
        for (int i=1; i<n; i++) {
            if(obstacleGrid[0][i] == 0)
                tab[0][i] = tab[0][i-1];
            else
                tab[0][i] = 0;
        }
        
        for (int i=1; i<m; i++) {
            if (obstacleGrid[i][0] == 0)
                tab[i][0] = tab[i-1][0];
            else
                tab[i][0] = 0;
        }
        
        
        for (int i=1; i<m; i++)
            for (int j=1; j<n; j++) {
                if (obstacleGrid[i][j] == 0)
                    tab[i][j] = tab[i][j-1] + tab[i-1][j];
                else
                    tab[i][j] = 0;
            }
            
        return tab[m-1][n-1];
        
    }
};