uniquePathsII.java

// Source : https://oj.leetcode.com/problems/unique-paths-ii/
// Inspired by : http://www.jiuzhang.com/solutions/unique-paths/
// Author : Lei Cao
// Date   : 2015-10-11

/**********************************************************************************
 *
 * 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.
 *
 **********************************************************************************/

package dynamicProgramming.uniquePaths;

public class uniquePathsII {
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if (obstacleGrid.length == 0 || obstacleGrid[0].length ==0) {
            return 0;
        }
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        // write your code here
        int[][] matrix = new int[m][n];
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] != 1) {
                matrix[i][0] = 1;
            } else {
                break;
            }
        }
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] != 1) {
                matrix[0][i] = 1;
            } else {
                break;
            }
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    matrix[i][j] = 0;
                } else {
                    matrix[i][j] = matrix[i-1][j] + matrix[i][j-1];
                }
            }
        }
        return matrix[m-1][n-1];
    }
}