# Unique Paths II

### 描述​

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 3 × 3 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.

### 备忘录法​

// Unique Paths II// 深搜 + 缓存，即备忘录法public class Solution {    public int uniquePathsWithObstacles(int[][] obstacleGrid) {        final int m = obstacleGrid.length;        final int n = obstacleGrid[0].length;        if (obstacleGrid[0][0] != 0 ||                obstacleGrid[m - 1][n - 1] != 0) return 0;        f = new int[m][n];        f[0][0] = obstacleGrid[0][0] != 0 ? 0 : 1;        return dfs(obstacleGrid, m - 1, n - 1);    }    // @return 从 (0, 0) 到 (x, y) 的路径总数    int dfs(int[][] obstacleGrid, int x, int y) {        if (x < 0 || y < 0) return 0; // 数据非法，终止条件        // (x,y)是障碍        if (obstacleGrid[x][y] != 0) return 0;        if (x == 0 && y == 0) return f[0][0]; // 回到起点，收敛条件        if (f[x][y] > 0) {            return f[x][y];        } else {            return f[x][y] = dfs(obstacleGrid, x - 1, y) +                    dfs(obstacleGrid, x, y - 1);        }    }    private int[][] f;  // 缓存}

### 动规​

// Unique Paths II// 动规，滚动数组// 时间复杂度 O(n^2)，空间复杂度 O(n)public class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {final int m = obstacleGrid.length;final int n = obstacleGrid[0].length;if (obstacleGrid[0][0] != 0 ||obstacleGrid[m-1][n-1] != 0) return 0;        int[] f = new int[n];        f[0] = obstacleGrid[0][0] != 0 ? 0 : 1;        for (int i = 0; i < m; i++) {            f[0] = f[0] == 0 ? 0 : (obstacleGrid[i][0] != 0 ? 0 : 1);            for (int j = 1; j < n; j++)                f[j] = obstacleGrid[i][j] != 0 ? 0 : (f[j] + f[j - 1]);        }        return f[n - 1];    }}