63. Unique Paths II(有障碍的路径 动态规划)

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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.

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int rows = obstacleGrid.length;
        int cols = obstacleGrid[0].length;
        int[][] dp = new int[rows][cols];
        
        for(int i = 0; i < rows;i++){
            for (int j = 0; j < cols ;j++){
                if(obstacleGrid[i][j]==1)
                    dp[i][j] = 0;
                else{
                    if (i==0&& j==0) 
                        dp[i][j] = 1;
                    else if (i==0) 
                        dp[i][j] = dp[i][j-1]; //边界 

                    else if (j == 0)
                        dp[i][j] = dp[i-1][j];
                    else 
                        dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
        }
        return dp[rows-1][cols-1];
    }
}

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int rows = obstacleGrid.length;
        int cols = obstacleGrid[0].length;
        for(int i = 0; i < rows;i++){
            for (int j = 0; j < cols ;j++){
                if(obstacleGrid[i][j]==1)
                    obstacleGrid[i][j] = 0;
                else if (i==0&& j==0) 
                    obstacleGrid[i][j] = 1;
                else if (i==0) 
                    obstacleGrid[i][j] = obstacleGrid[i][j-1]*1; //边界，没有路径了，要么是0，要么是1
                
                else if (j == 0)
                    obstacleGrid[i][j] = obstacleGrid[i-1][j]*1;
                else 
                    obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
            }
        }
        return obstacleGrid[rows-1][cols-1];
    }
}