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.

方法一：与I一样，动态规划。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = obstacleGrid.size();
        int m = obstacleGrid[0].size();
        
        if(obstacleGrid[0][0] == 1 || obstacleGrid[n-1][m-1] == 1) return 0;
        
        obstacleGrid[0][0] = 1;
        
        for(int i = 1;i<n;i++)
        {
            obstacleGrid[i][0] = (obstacleGrid[i][0] == 0)?obstacleGrid[i-1][0]:0;
        }
        
        for(int i = 1;i<m;i++)
        {
            obstacleGrid[0][i] = (obstacleGrid[0][i] == 0)?obstacleGrid[0][i-1]:0;
        }
        
        for(int i = 1;i<n;i++)
        {
            for(int j = 1;j < m;j++)
            {           
                obstacleGrid[i][j] = (obstacleGrid[i][j] == 0)?(obstacleGrid[i-1][j] + obstacleGrid[i][j-1]):0;
            }
        }
        
        return obstacleGrid[n-1][m-1];
    }
};

28 milli secs

时间复杂度O(n^2)，空间O(n^2)，但在这道题中没有extra space

另一种写法:

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = obstacleGrid.size();
        int m = obstacleGrid[0].size();
        
        if(obstacleGrid[0][0] == 1 || obstacleGrid[n-1][m-1] == 1) return 0;
        
        vector<int> dp(m+1,0);
        int j = m-1;
        while(j>=0 && obstacleGrid[n-1][j] == 0)
            dp[j--] = 1;
        
        j = n-1;
        while(--j >= 0)
        {
            for(int k = m-1;k>=0;k--)
                dp[k] = (obstacleGrid[j][k] == 0)?dp[k] + dp[k+1]:0;
        }
        
        return dp[0];
    }
};

20 milli secs