问题描写叙述：

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.

Note: m and n will be at most 100.

基本思路：

此题是前篇 的变形。

增加了障碍格子。可是思路与前篇基本一致。仅仅是在前篇的基础上考虑了障碍格子。

变化的地方有两个：

1. 初始化第一行第一列时考虑有障碍的影响。

   一旦行或列中出现了障碍格子。后面的格子都不可达。

2. 在更新到达（i，j）格子时。要注意查看（i。j-1）和（i-1,j）是否是障碍格子，并分别处理。

代码：

int uniquePathsWithObstacles(vector
     
      
        > &obstacleGrid) {  //c++        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        int array[m][n];                  //init           bool haveObs = false;        for(int i = 0; i < n; i++){              if(obstacleGrid[0][i] == 0 && !haveObs){                array[0][i] = 1;                  continue;               }            else if(obstacleGrid[0][i] == 1){                haveObs = true;            }            array[0][i] = 0;        }                haveObs = false;        for(int i = 0; i < m; i++) {             if(obstacleGrid[i][0] == 0 && !haveObs){                array[i][0] = 1;                continue;               }            else if(obstacleGrid[i][0 ==1]){                haveObs = true;                }             array[i][0] = 0;        }                      for(int i = 1; i < m; i++)              for(int j = 1; j < n; j++){                 array[i][j] = 0;                if(obstacleGrid[i][j] == 1)                    continue;                if(obstacleGrid[i-1][j] == 0)                    array[i][j] += array[i-1][j];                if(obstacleGrid[i][j-1] == 0)                    array[i][j] +=array[i][j-1];              }          return array[m-1][n-1];      }