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.

Solution1 DP Space(m * n)

1ms 19.48%

public class Solution {
public int uniquePathsWithObstacles(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] path = new int[m]
;
for (int i = 0; i < m; i++) {
if (grid[i][0] != 1) {
path[i][0] = 1;
} else {
break;
}
}

for (int i = 0; i < n; i++) {
if (grid[0][i] != 1) {
path[0][i] = 1;
} else {
break;
}
}

for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (grid[i][j] == 1) {
path[i][j] = 0;
}else {
path[i][j] = path[i - 1][j] + path[i][j - 1];
}
}
}
return path[m - 1][n - 1];
}
}

Solution 2 DP O(n) space

1ms 19.48%

public static int uniquePathsWithObstacles2(int[][] obstacleGrid) {
int width = obstacleGrid[0].length;
int[] dp = new int[width];
dp[0] = 1;
for (int[] row : obstacleGrid) {
for (int j = 0; j < width; j++) {
if (row[j] == 1)
dp[j] = 0;
else if (j > 0)
dp[j] += dp[j - 1];// don't miss +
}
}
return dp[width - 1];
}Solution 3 DP O(1) space

public static int uniquePathsWithObstacles3(int[][] obstacleGrid) {
if(obstacleGrid.length == 0) return 0;

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];// For row 0, if there are no paths to left cell, then its 0,else 1
else if(j == 0)
obstacleGrid[i][j] = obstacleGrid[i - 1][j];// For col 0, if there are no paths to upper cell, then its 0,else 1
else
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
}
}

return obstacleGrid[rows - 1][cols - 1];
}