Unique Paths II

Unique Paths II

问题：

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.

思路：

　　简单的动态规划

我的代码：

public class Solution {

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {

        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;

        int m = obstacleGrid.length;

        int n = obstacleGrid[0].length;

        if(obstacleGrid[0][0] == 1)

            obstacleGrid[0][0] = 0;

        else

            obstacleGrid[0][0] = 1;

        for(int k = 1; k < m; k++)

        {

            if(obstacleGrid[k][0] != 1)

                obstacleGrid[k][0] = obstacleGrid[k-1][0];

            else

                obstacleGrid[k][0] = 0;

        }

        for(int k = 1; k < n; k++)

        {

            if(obstacleGrid[0][k] != 1)

                obstacleGrid[0][k] = obstacleGrid[0][k-1];

            else

                obstacleGrid[0][k] = 0;

        }

        for(int i = 1; i < m; i++)

        {

            for(int j = 1; j < n; j++)

            {

                if(obstacleGrid[i][j] == 1)

                {

                    obstacleGrid[i][j] = 0;

                }

                else

                {

                    obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];

                }

            }

        }

        return obstacleGrid[m-1][n-1];

    }

}

View Code

别人代码：

public class Solution {

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {

        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {

            return 0;

        }

        

        int n = obstacleGrid.length;

        int m = obstacleGrid[0].length;

        int[][] paths = new int[n][m];

        

        for (int i = 0; i < n; i++) {

            if (obstacleGrid[i][0] != 1) {

                paths[i][0] = 1;

            } else {

                break;

            }

        }

        

        for (int i = 0; i < m; i++) {

            if (obstacleGrid[0][i] != 1) {

                paths[0][i] = 1; 

            } else {

                break;

            }

        }

        

        for (int i = 1; i < n; i++) {

            for (int j = 1; j < m; j++) {

                if (obstacleGrid[i][j] != 1) {

                    paths[i][j] = paths[i - 1][j] + paths[i][j - 1];

                } else {

                    paths[i][j] = 0;

                }

            }

        }

        

        return paths[n - 1][m - 1];

    }

}

View Code

学习之处：

• 对于第一行的corner case 需要考虑的较多，所以AC了几次还是没过去，看别人代码里面 对于第一行或者第一列中那个break用的真是恰到好处，节约了时间，解决了corner case，学习一下