[leetcode-63-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.

思路：

类似于

leetcode-62-Unique Paths中提到的方法，仅需判断当前格子是否是障碍即可。

如果是第一行或者第一列的障碍，那么将它之后所有格子置为0。

如果不是第一行或者第一列的话，仅需将它置为0。

int uniquePathsWithObstacles(vector<vector<int>>& grid)
    {
        //类似没有障碍的方法：将有障碍的地方置为0；
        if (grid.empty())return 0;
        int row = grid.size();
        int col = grid[0].size();
        if (grid[0][0] == 1)return 0;
        
        for (int i = 0; i < col;i++)//处理第一行
        {
            if (grid[0][i] == 0)grid[0][i] = 1;            
            else if (grid[0][i] == 1)
            {
                for (int j = i; j < col; j++)
                    grid[0][j] = 0;//i之后都置为0;
                i = col;
            }                
        }
        for (int i = 1; i < row; i++)//处理第一列
        {
            if (grid[i][0] == 0)grid[i][0] = 1;
            else if (grid[i][0] == 1)
            {
                for (int j = i; j < row; j++)
                    grid[j][0] = 0;//i之后都置为0;
                i = row;
            }
        }
        for (int i = 1; i < row; i++)
        {
            for (int j = 1; j < col; j++)
            {
                if (grid[i][j] == 0)//不是障碍
                {                                
                    grid[i][j] = grid[i - 1][j] + grid[i][j - 1];                    
                }
                else if(grid[i][j] == 1)grid[i][j] = 0;
            }
        }
        return grid[row - 1][col - 1];
    }