leetcode oj java Unique Paths II

一、题目描述：

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

.

二、解决方法：

Unique Paths 的解决方法在上一篇文章中写出（动态规划） http://blog.csdn.net/u011060119/article/details/53764150

这次增加了障碍物，我们需要在障碍物的部分把路径数目置0，（障碍物垂直向下和水平向右都需要置0）

三、代码：

package T12;

/**
* @author 作者 : xcy
* @version 创建时间：2016年12月20日 下午4:03:08
* 类说明
*/
public class t63 {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] obstacleGrid = { { 0, 0 }, { 1, 1 }, { 0, 0 } };
System.out.println(uniquePathsWithObstacles(obstacleGrid));

}

public static int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
if (obstacleGrid[0][0] == 1) {
return 0;
}
int[][] re = new int[m]
;
for (int i = 0; i < m; i++) {
if (obstacleGrid[i][0] == 0) {
re[i][0] = 1;
} else {
break;
}
}
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] == 0) {
re[0][j] = 1;
} else {
break;
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 0) {
re[i][j] = re[i - 1][j] + re[i][j - 1];
}
}
}
return re[m - 1][n - 1];
}

}