unique PathsII：带障碍寻径

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 as1and0respectively 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 is2.

Note: m and n will be at most 100.

思路：这题不用转载了，因为做完寻径1后，此题和它是一个思路，仍用动态规划思路，当遇到障碍时直接让词典可走路径数为零。

注意，由于可能从开始位置就不能走，所以一维数组第一位也要参与更新，即不在默认人为一定存在以条沿左下边路直奔终点的路径！（这个位置真的好容易错啊）

最下面是递归求解，时间超时。

public int uniquePathsWithObstacles(int[][] obstacleGrid) {        int m = obstacleGrid.length;        if(m==0) return 0;        int n = obstacleGrid[0].length;               return dp(obstacleGrid,m,n);                    }     public int dp(int[][] o,int m,int n){   int[] d = new int[n];   if(o[m-1][n-1]!=1) d[0] = 1;   for(int i=0;i<m;i++){   for(int j=0;j<n;j++){   if(o[i][j]==1){   d[j]=0;   }else{                   if(j>0){                       d[j]+=d[j-1];                   }      }   }   }   return d[n-1];   }   public int dfs(int[][] o,int x,int y,int m,int n){    System.out.println("x,y:"+x+y);    if(x>m-1||y>n-1){            return 0;        }        if(o[x][y]==1) return 0;        if(x==m-1&&y==n-1){            return 1;        }                int r1=dfs(o,x+1,y,m,n);        int r2=dfs(o,x,y+1,m,n);        return r1+r2;    }