// 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]; } }