[LeetCode] 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.

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 Array Dynamic Programming

 

思路：在Unique Paths的基础上，加上obstacleGrid[i-1][j-1]==0 时，f[i][j]=0,另外，注意红色部分的特殊处理。

class Solution {
public:
    
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
        {   
            int m = obstacleGrid.size();
            int n = obstacleGrid[0].size();

            vector<int> row(n + 1, 0); 
            vector<vector<int> > f(m + 1, row);

            if(obstacleGrid[0][0] == 1)
                return 0;
            f[1][1] = 1;

            for(int i = 1; i <= m; i ++) 
            {   
                for(int j = 1; j <= n; j ++) 
                {   
                    if(i == 1 && j == 1)
                        continue;
                    if(obstacleGrid[i-1][j-1] == 1)
                        f[i][j] = 0;
                    else
                        f[i][j] = f[i-1][j] + f[i][j-1];
                }   
            }   

            return f[m][n];
        }    
    
};

 思路二：上门的思路改成滚动数组

class Solution {
public:
    
        int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
        {
            int m = obstacleGrid.size();
            int n = obstacleGrid[0].size();

            if(obstacleGrid[0][0] == 1)
                return 0;
            vector<int> f(n + 1, 0);

            f[1] = 1;

            for(int i = 1; i <= m; i ++)
            {
                for(int j = 1; j <= n; j ++)
                {
                    if(i == 1 && j == 1)
                        continue;
                    if(obstacleGrid[i-1][j-1] == 1)
                        f[j] = 0;
                    else
                        f[j] = f[j] + f[j-1];
                }
            }

            return f[n];
        }  
    
};