leetcode oj java Unique Paths II

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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][n];        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];    }}

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