/** * Source : https://oj.leetcode.com/problems/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 UniquePath2 {    /**     * 依然使用动态规划     * 注意障碍，障碍在边上和中间     *     * @param maze     * @return     */    public int finAllUniquePaths (int[][] maze) {        if (maze.length <= 0 || maze[0].length <= 0) {            return 0;        }        int max = 0;        for (int i = 0; i < maze.length; i++) {            for (int j = 0; j < maze[0].length; j++) {                if (maze[i][j] == 1) {                    // 障碍处为0                    max = maze[i][j] = 0;                } else {                    if (i > 0 && j > 0) {                        max = maze[i][j] = maze[i-1][j] + maze[i][j-1];                    } else if (i > 0) {                        // 第一列不一定是1                        max = maze[i][j] = maze[i-1][j];                    } else if (j > 0) {                        // 第一行不一定是1                        max = maze[i][j] = maze[i][j-1];                    } else {                        // 第一个是1                        max = maze[i][j] = 1;                    }                }            }        }        return max;    }    public static void main(String[] args) {        UniquePath2 uniquePaths = new UniquePath2();        int[][] arr = new int[][]{                {0,1},                {0,0}        };        int[][] arr1 = new int[][]{                {0,1,0},                {0,0,0}        };        int[][] arr2 = new int[][]{                {0,1,0},                {0,1,0},                {0,0,0}        };        int[][] arr3 = new int[][]{                {0,0,0},                {0,1,0},                {0,0,0}        };        int[][] arr4 = new int[][]{                {0,0,0,0,0,0,0},                {0,1,0,0,0,0,0},                {0,0,0,0,0,0,0}        };        System.out.println(uniquePaths.finAllUniquePaths(arr));        System.out.println(uniquePaths.finAllUniquePaths(arr1));        System.out.println(uniquePaths.finAllUniquePaths(arr2));        System.out.println(uniquePaths.finAllUniquePaths(arr3));        System.out.println(uniquePaths.finAllUniquePaths(arr4));    }}