// Author: Li Long, [email protected]

// Date: Apr 17, 2014

// Source: http://oj.leetcode.com/problems/unique-paths-ii/

// Analysis: http://blog.csdn.net/lilong_dream/article/details/19931301

// 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 UniquePathsII {

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

int m = obstacleGrid.length;

// if (m == 0) {

// return 0;

// }

int n = obstacleGrid[0].length;

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

return 0;

} else if (m == 1 && n == 1) {

return 1;

}

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

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

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

while (i < m) {

paths[i][0] = 0;

++i;

}

break;

} else {

paths[i][0] = 1;

}

}

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

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

while (j < n) {

paths[0][j] = 0;

++j;

}

break;

} else {

paths[0][j] = 1;

}

}

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

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

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

paths[i][j] = 0;

} else {

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

}

}

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

}

public static void main(String[] args) {

int[][] obstacle = new int[][] { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };

UniquePathsII slt = new UniquePathsII();

int result = slt.uniquePathsWithObstacles(obstacle);

System.out.println(result);

}

}