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Unique Paths II

2024-07-17 03:49:28   219人阅读

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.

 1 class Solution { 2 public: 3     int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { 4         int m = obstacleGrid.size(); 5         if(m == 0) return 0; 6         int n = obstacleGrid[0].size(); 7          8         vector<int> row(n,0); 9         vector<vector<int>> record(m,row);10         11         bool obstacle = false;12         for(int i = n-1; i >= 0; i--){13             if(obstacle)14                 record[m-1][i] = 0;15             else if(obstacleGrid[m-1][i] == 1){16                 obstacle = true;17                 record[m-1][i] = 0;18             }else record[m-1][i] = 1;19         }20         obstacle = false;21         for(int i = m-1; i >= 0; i--){22             if(obstacle)23                 record[i][n-1] = 0;24             else if(obstacleGrid[i][n-1] == 1){25                 obstacle = true;26                 record[i][n-1] = 0;27             }else record[i][n-1] = 1;28             29         }30         for(int i = m-2; i >= 0; i--)31             for(int j = n-2; j >= 0; j--){32                 if(obstacleGrid[i][j] == 1) record[i][j] = 0;33                 else record[i][j] = record[i+1][j] + record[i][j+1];34             }35         return record[0][0];36     }37 };

动态规划，时间复杂度O(m*n), 空间复杂度O(m*n)。


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