*Unique paths II

题目：

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.

解法一：我的解法

    public int uniquePathsWithObstacles(int[][] obstacleGrid) 
    {
        int m=obstacleGrid.length;
        //if(m==0) return 1;
        int n=obstacleGrid[0].length;
        int [][] grids = new int [m+1][n+1];
        grids[m][n-1]=1;
        for(int i=m-1;i>=0;i--)
        {
            for (int j=n-1;j>=0;j--)
            {
                grids[i][j]=(obstacleGrid[i][j]==1) ? 0:grids[i+1][j]+grids[i][j+1];
            }
        }
        
        return grids[0][0];
        
    }

解法二：九章算法的解法

    public int uniquePathsWithObstacles(int[][] obstacleGrid) 
    {
        int m=obstacleGrid.length;
        //if(m==0) return 1;
        int n=obstacleGrid[0].length;
        int [][] grids = new int [m][n];
        for (int i = 0; i < m; i++) 
        {
            if (obstacleGrid[i][0] != 1) {
                grids[i][0] = 1;
            } else {
                //grids[i][0] = 0;
                break;   //第一列！错过了就不会再回到第一列！只能往右边和下边走。
            }
        }
        
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] != 1) {
                grids[0][i] = 1; 
            } else {
                //grids[0][i] = 0;
                break;   //第一行！错过了就不会回到第一行。
            }
        }
        
        
        
        for(int i=1;i<m;i++)
        {
            for (int j=1;j<n;j++)
            {
                grids[i][j]=(obstacleGrid[i][j]==1) ? 0:grids[i-1][j]+grids[i][j-1];
            }
        }
        
        return grids[m-1][n-1];
        
    }