63. Unique Paths II

题目：

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

 

链接：  http://leetcode.com/problems/unique-paths-ii/

题解：

也是DP问题，Unique Path一样可以in place解决。要点是在设置第一行和第一列碰到obstacle的时候，要将其以及之后的所有值设置为零，因为没有路径可以达到。之后在DP扫描矩阵的时候，也要讲obstacle所在的位置清零。

Time Complexity - O(m * n)， Space Complexity - O(1)。

public class Solution {

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

        int rowNum = obstacleGrid.length, colNum = obstacleGrid[0].length;

        if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0][0] == 1){

            return 0;

        }

        

        for(int row = 0; row < rowNum; row ++){

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

                obstacleGrid[row][0] = 1;

            else if (obstacleGrid[row][0] == 1){              //if find obstacle, set all [row,0] below obstacle to 0

                for(int tempRow = row; tempRow < rowNum; tempRow ++)

                    obstacleGrid[tempRow][0] = 0;

                break;

            }

        }

        

         for(int col = 1; col < colNum; col ++){

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

                obstacleGrid[0][col] = 1;

            else if (obstacleGrid[0][col] == 1){            // //if find obstacle, set all [0,col] one the right of obstacle to 0

                for(int tempCol = col; tempCol < colNum; tempCol ++)

                    obstacleGrid[0][tempCol] = 0;

                break;

            }

        }

        

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

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

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

                    obstacleGrid[i][j] = 0;

                else

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

            }

        }

        

        return obstacleGrid[rowNum - 1][colNum - 1];

    }

}

 

测试：