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.

细节多注意

这类vector的长度求法

obstacleGrid.size();

obstacleGrid[0].size();

用length()报错

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) 
    {
        if(obstacleGrid[0][0]==1)return 0;
        int i,j;
        int row = obstacleGrid.size();  
        int column = obstacleGrid[0].size();  
        int dp[row+1][column+1];
        dp[0][0]=1;
        for(i=1;i<column;i++)
        {
            if(obstacleGrid[0][i]==0&&dp[0][i-1]==1)
                dp[0][i]=1;
            else 
                dp[0][i]=0;
        }
        for(i=1;i<row;i++)
        {
            if(obstacleGrid[i][0]==0&&dp[i-1][0]==1)
                dp[i][0]=1;
            else 
                dp[i][0]=0;
        }
        for(i=1;i<row;i++)
        for(j=1;j<column;j++)
        {
            if(obstacleGrid[i][j]==1)
            {
                dp[i][j] = 0;
                continue;
            }
            if(obstacleGrid[i-1][j]==1&&obstacleGrid[i][j-1]==1)
            {
                dp[i][j] = 0;
                continue;
            }
                
            if(obstacleGrid[i-1][j]==1&&obstacleGrid[i][j-1]==0)
                dp[i][j] = dp[i][j-1];
            if(obstacleGrid[i-1][j]==0&&obstacleGrid[i][j-1]==1)
                dp[i][j] = dp[i-1][j];
            if(obstacleGrid[i-1][j]==0&&obstacleGrid[i][j-1]==0)
                dp[i][j] = dp[i][j-1] + dp[i-1][j];
                
        }
        return dp[row-1][column-1];
    }
};