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.

思路： Using a integer array to storage the path. If the path is a obstacle, make it 0, else equals to the sum of left and up;

易错点： 初始化时候，横竖第一行， 只要有一个障碍物， 后面就全设置成0， 因为被阻了。 还有要判断这个不是障碍物时候才能填充， 防止 最后一个终点是障碍。

public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        int m = obstacleGrid.length;        int n = obstacleGrid[0].length;        int[][] path = new int[m][n];        for(int i = 0; i < m; i++){            if(obstacleGrid[i][0] == 1)//--                break;            path[i][0] = 1;        }        for(int j = 0; j < n; j++){            if(obstacleGrid[0][j] == 1)//--                break;            path[0][j] = 1;        }        for(int i = 1; i < m; i++){            for(int j = 1; j < n; j++){                if(obstacleGrid[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];    }}