A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
Now consider if some obstacles are added to the grids. How many unique paths would there be?

一個機器人從一個mxn的網格左上角出發，只能向下或向右移動，最終要到達右下角，問可能有多少條不同的路徑。網格中有一些障礙物由1表示，見例一：

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
有兩種途徑到達右下角:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
 

動態規劃：
使用dp陣列來儲存每一點可能的路徑數。我最開始想的是用三維陣列存，形成思維定勢了，但是太慢了。後來發現只能往右邊和下邊走，意味著不能返回，因此二維陣列就夠了。

class Solution {
    public static int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int rows = obstacleGrid.length;
        int cols = obstacleGrid[0].length;
        int[][] dp = new int[rows][cols]
 
;
        if (obstacleGrid[0][0] == 0) dp[0][0] = 1;
        for (int i = 1; i < rows; i++) {
            dp[i][0] = obstacleGrid[i][0] == 0 ? dp[i - 1][0] : 0;
        }
        for (int i = 1; i < cols; i++) {
            dp[0][i] = obstacleGrid[0][i] == 0 ? dp[0][i - 1] : 0;
        }
        for (int i = 1; i < rows; i++) {
            for (int j = 1; j < cols; j++) {
                dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[rows - 1][cols - 1];
    }
}

後來又看了一下推薦解法，空間複雜度可以達到O（1），是用的原陣列存的dp陣列的內容。

程式碼：

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

        int R = obstacleGrid.length;
        int C = obstacleGrid[0].length;

        // If the starting cell has an obstacle, then simply return as there would be
        // no paths to the destination.
        if (obstacleGrid[0][0] == 1) {
            return 0;
        }

        // Number of ways of reaching the starting cell = 1.
        obstacleGrid[0][0] = 1;

        // Filling the values for the first column
        for (int i = 1; i < R; i++) {
            obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i - 1][0] == 1) ? 1 : 0;
        }

        // Filling the values for the first row
        for (int i = 1; i < C; i++) {
            obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i - 1] == 1) ? 1 : 0;
        }

        // Starting from cell(1,1) fill up the values
        // No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
        // i.e. From above and left.
        for (int i = 1; i < R; i++) {
            for (int j = 1; j < C; j++) {
                if (obstacleGrid[i][j] == 0) {
                    obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
                } else {
                    obstacleGrid[i][j] = 0;
                }
            }
        }

        // Return value stored in rightmost bottommost cell. That is the destination.
        return obstacleGrid[R - 1][C - 1];
    }
}