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.
类似于Unique Paths的分析方法，用动态规划，初始化第一行和第一列，遇到1，后面和下面的都设成0，然后就是每个格加左边的和上面的，碰到1，这个格设成0。代码如下：
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid.length == 0 || obstacleGrid[0].length == 0 || obstacleGrid[0][0] == 1) {
return 0;
}
int init = 1;
for (int i = 0; i < obstacleGrid.length; i ++) {
if (obstacleGrid[i][0] == 1) {
init = 0;
}
obstacleGrid[i][0] = init;
}
init = 1;
for (int i = 1; i < obstacleGrid[0].length; i ++) {
if (obstacleGrid[0][i] == 1) {
init = 0;
}
obstacleGrid[0][i] = init;
}
for (int i = 1; i < obstacleGrid.length; i ++) {
for (int j = 1; j < obstacleGrid[0].length; j ++) {
if (obstacleGrid[i][j] == 1) {
obstacleGrid[i][j] = 0;
} else {
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
}
}
}
return obstacleGrid[obstacleGrid.length - 1][obstacleGrid[0].length - 1];
}
}也可以用一维数组存放步数，代码如下：
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int width = obstacleGrid[0].length;
int[] dp = new int[width];
dp[0] = 1;
for (int i = 0; i < obstacleGrid.length; i ++) {
for (int j = 0; j < width; j ++) {
if (obstacleGrid[i][j] == 1) {
dp[j] = 0;
} else if (j > 0) {
dp[j] = dp[j] + dp[j - 1];
}
}
}
return dp[width - 1];
}
}