Unique Paths II Java Solutions

Follow up for "Unique Paths":

Now consider if some obstacles is added to the grids. How many unique paths would there be?

An obstacle and empty space are marked as and respectively in the 1 0 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 .

The problem and

Unique Paths Java Solutions

Compared to a more barrier setting, as long as the DP initialization of the barrier special settings, in the middle DP process plus judgment can be.

1  Public classSolution {2      Public intUniquepathswithobstacles (int[] obstaclegrid) {3         if(Obstaclegrid = =NULL|| Obstaclegrid.length = = 0 | | Obstaclegrid[0].length = = 0)return0;4         intm = obstaclegrid.length, n = obstaclegrid[0].length;5         int[] DP =New int[m][n];6         BooleanFlag =true;7          for(inti = 0; I < m; i++){8             if(obstaclegrid[i][0] = = 0 && Flag = =true){9Dp[i][0] = 1;Ten}Else{ OneFlag =false; ADp[i][0] = 0; -             } -         } theFlag =true; -          for(inti = 0; I < n; i++){ -             if(Obstaclegrid[0][i] = = 0 && Flag = =true){ -Dp[0][i] = 1; +}Else{ -Flag =false; +Dp[0][i] = 0; A             } at         } -          for(inti = 1; I < m; i++){ -              for(intj = 1; J < N; J + +){ -                 if(Obstaclegrid[i][j] = = 0) -DP[I][J] = Dp[i-1][j] + dp[i][j-1]; -                 Else inDP[I][J] = 0; -             } to         } +         returnDp[m-1][n-1]; -     } the}

Optimization for one-dimensional array solution:

1  Public classSolution {2      Public intUniquepathswithobstacles (int[] obstaclegrid) {3         if(Obstaclegrid = =NULL|| Obstaclegrid.length = = 0 | | Obstaclegrid[0].length = = 0)return0;4         int[] DP =New int[Obstaclegrid[0].length];5Dp[0] = 1;6          for(inti = 0; i < obstaclegrid.length; i++)7              for(intj = 0; J < Obstaclegrid[0].length; J + +)8                 if(Obstaclegrid[i][j] = = 1) dp[j] = 0;9                 Else if(J > 0) dp[j] + + dp[j-1];Ten         returnDp[obstaclegrid[0].length-1]; One     } A}

DP[J] On the previous line,

Dp[j-1] left of current position

It can therefore be optimized for one-dimensional arrays.

Unique Paths II Java Solutions