Unique Paths II

原题：

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.

返回从给定数组左上点走到右下点路径数。其中二维数组值为1的是障碍，不能走。

思考过程：

还是动态规划的思路。其中把障碍点的路径数设为1.边界点的路径数是沿着边界它前一个点的路径数。

AC代码：

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int n = obstacleGrid.length,m = obstacleGrid[0].length;
if (obstacleGrid[n-1][m-1] == 1 || obstacleGrid[0][0] == 1) return 0;//起点或终点有障碍的，路径数是0
int[][] matrix = new int
[m];
for (int i = 0;i < n;i++)
for (int j = 0;j < m;j++){
int x = 0;
if (i == 0 && j == 0)
x = 1;//起点路径数设为1)
else if (obstacleGrid[i][j] == 1)
x = 0;//障碍处路径数设为0
else if (i == 0) x = matrix[i][j-1];//边界点的路径数等于沿着边界前一个点的路径数
else if (j == 0) x = matrix[i-1][j];
else x = matrix[i - 1][j] + matrix[i][j - 1];
matrix[i][j] = x;
}
return matrix[n - 1][m - 1];
}