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.

public class Solution {
    public int uniquePathsWithObstacles (int[][] obstacleGrid) {
        //开始应该判断一些 obstacleGrid 是不是为 null 或者为不为空
        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;
        int[][] map = new int[row][col];
        /**
         * 记得考虑两种情况
         * 1 当要到达的点本身有 obstacle， 则直接返回0
         * 2 要记得考虑边界值，即最后一列和最后一行的算法跟其他的不一样
         */
        if (obstacleGrid[row - 1][col - 1] == 1){
            return 0;
        }else {
            map[row - 1][col - 1] = 1;
        
        for (int i = col - 2; i >= 0; i--) {
                if (obstacleGrid[row-1][i] == 1) {
                    //其实直接break 就好，因为数组初始化时，每个变量都默认为0了
                    map[row-1][i] = 0;
                }else {
                    map[row-1][i] = map[row-1][i+1];
                }
        }
        
        for (int i = row - 2; i >= 0; i--) {
                if (obstacleGrid[i][col-1] == 1) {
                    map[i][col-1] = 0;
                }else {
                    map[i][col-1] = map[i+1][col-1];
                }
        }

        for (int i = row - 2; i >= 0; i--) {
            for (int j = col - 2; j >= 0; j--) {
                if (obstacleGrid[i][j] == 1) {
                    map[i][j] = 0;
                }else {
                    map[i][j] = map[i][j+1] + map[i+1][j];
                }
            }
        }
        
        return map[0][0];
        }   

别人的答案

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];
    }
}

这是第二次写的程序，这里比上面方法效率低的一点是 当初始化第一行和第一列的时候，如果某处obstacleGrid 的点为1， 则result 这个点以后全都为0即可，即直接break 结束循环即可（看上面程序）

public class Solution {
    public int uniquePathsWithObstacles (int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {
            return 0;
        }
        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;
        int[][] result = new int[row][col];
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        result[0][0] = 1;
        for (int i = 1; i < col; i++) {
            result[0][i] = obstacleGrid[0][i] == 1 ? 0 : result[0][i - 1];
        }
        for (int i = 1; i < row; i++) {
            result[i][0] = obstacleGrid[i][0] == 1 ? 0 : result[i - 1][0];
        }
        for (int i = 1; i < row; i++) {
            for (int j = 1; j < col; j++) {
                result[i][j] = obstacleGrid[i][j] == 1 ? 0 : result[i - 1][j] + result[i][j - 1];
            }
        }
        return result[row - 1][col - 1];
    }
}

本来想optimize 的，结果又漏掉corner case

public class Solution {
    public int uniquePathsWithObstacles (int[][] obstacleGrid) {
        if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) {
            return 0;
        }
        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;
        int[][] result = new int[row][col];
        for (int i = 0; i < col; i++) {
            if (obstacleGrid[0][i] != 1) {
                result[0][i] = 1;
            } else {
                break;
            }
        }
        //for (int i = 1; i < row; i++) 
        // 如果写成 i 从1 开始，则对于obstacleGrid[0][0] = 1 的这种情况考虑出错
        //{{1, 0}} test case return 1，wrong answer 
        for (int i = 0; i < row; i++) {
            if (obstacleGrid[i][0] != 1) {
                result[i][0] = 1;
            } else {
                break;
            }
        }
        for (int i = 1; i < row; i++) {
            for (int j = 1; j < col; j++) {
                result[i][j] = obstacleGrid[i][j] == 1 ? 0 : result[i - 1][j] + result[i][j - 1];
            }
        }
        return result[row - 1][col - 1];
    }
}