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63. 不同路径 II

题目描述

给定一个 m x n 的整数数组 grid。一个机器人初始位于 左上角（即 grid[0][0]）。机器人尝试移动到 右下角（即 grid[m - 1][n - 1]）。机器人每次只能向下或者向右移动一步。

网格中的障碍物和空位置分别用 1 和 0 来表示。机器人的移动路径中不能包含 任何 有障碍物的方格。

返回机器人能够到达右下角的不同路径数量。

测试用例保证答案小于等于 2 * 109。

 

示例 1：

输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
输出：2
解释：3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

示例 2：

输入：obstacleGrid = [[0,1],[0,0]]
输出：1

 

提示：

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] 为 0 或 1

解法

方法一：记忆化搜索

我们设计一个函数 $dfs(i, j)$ 表示从网格 $(i, j)$ 到网格 $(m - 1, n - 1)$ 的路径数。其中 $m$ 和 $n$ 分别是网格的行数和列数。

函数 $dfs(i, j)$ 的执行过程如下：

  • 如果 $i \ge m$ 或者 $j \ge n$，或者 $obstacleGrid[i][j] = 1$，则路径数为 $0$；
  • 如果 $i = m - 1$ 且 $j = n - 1$，则路径数为 $1$；
  • 否则，路径数为 $dfs(i + 1, j) + dfs(i, j + 1)$。

为了避免重复计算，我们可以使用记忆化搜索的方法。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= m or j >= n or obstacleGrid[i][j]:
                return 0
            if i == m - 1 and j == n - 1:
                return 1
            return dfs(i + 1, j) + dfs(i, j + 1)

        m, n = len(obstacleGrid), len(obstacleGrid[0])
        return dfs(0, 0)

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class Solution {
    private Integer[][] f;
    private int[][] obstacleGrid;
    private int m;
    private int n;

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        m = obstacleGrid.length;
        n = obstacleGrid[0].length;
        this.obstacleGrid = obstacleGrid;
        f = new Integer[m][n];
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= m || j >= n || obstacleGrid[i][j] == 1) {
            return 0;
        }
        if (i == m - 1 && j == n - 1) {
            return 1;
        }
        if (f[i][j] == null) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    }
}

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        int f[m][n];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) {
            if (i >= m || j >= n || obstacleGrid[i][j]) {
                return 0;
            }
            if (i == m - 1 && j == n - 1) {
                return 1;
            }
            if (f[i][j] == -1) {
                f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
            }
            return f[i][j];
        };
        return dfs(0, 0);
    }
};

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func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m, n := len(obstacleGrid), len(obstacleGrid[0])
    f := make([][]int, m)
    for i := range f {
        f[i] = make([]int, n)
        for j := range f[i] {
            f[i][j] = -1
        }
    }
    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if i >= m || j >= n || obstacleGrid[i][j] == 1 {
            return 0
        }
        if i == m-1 && j == n-1 {
            return 1
        }
        if f[i][j] == -1 {
            f[i][j] = dfs(i+1, j) + dfs(i, j+1)
        }
        return f[i][j]
    }
    return dfs(0, 0)
}

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function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
            return 0;
        }
        if (i === m - 1 && j === n - 1) {
            return 1;
        }
        if (f[i][j] === -1) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    };
    return dfs(0, 0);
}

方法二：动态规划

我们定义 $f[i][j]$ 表示到达网格 $(i,j)$ 的路径数。

首先初始化 $f$ 第一列和第一行的所有值，然后遍历其它行和列，有两种情况：

  • 若 $obstacleGrid[i][j] = 1$，说明路径数为 $0$，那么 $f[i][j] = 0$；
  • 若 $obstacleGrid[i][j] = 0$，则 $f[i][j] = f[i - 1][j] + f[i][j - 1]$。

最后返回 $f[m - 1][n - 1]$ 即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        f = [[0] * n for _ in range(m)]
        for i in range(m):
            if obstacleGrid[i][0] == 1:
                break
            f[i][0] = 1
        for j in range(n):
            if obstacleGrid[0][j] == 1:
                break
            f[0][j] = 1
        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 0:
                    f[i][j] = f[i - 1][j] + f[i][j - 1]
        return f[-1][-1]

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class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[][] f = new int[m][n];
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
            f[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
            f[0][j] = 1;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 0) {
                    f[i][j] = f[i - 1][j] + f[i][j - 1];
                }
            }
        }
        return f[m - 1][n - 1];
    }
}

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class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<vector<int>> f(m, vector<int>(n));
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
            f[i][0] = 1;
        }
        for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
            f[0][j] = 1;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 0) {
                    f[i][j] = f[i - 1][j] + f[i][j - 1];
                }
            }
        }
        return f[m - 1][n - 1];
    }
};

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func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m, n := len(obstacleGrid), len(obstacleGrid[0])
    f := make([][]int, m)
    for i := 0; i < m; i++ {
        f[i] = make([]int, n)
    }
    for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
        f[i][0] = 1
    }
    for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {
        f[0][j] = 1
    }
    for i := 1; i < m; i++ {
        for j := 1; j < n; j++ {
            if obstacleGrid[i][j] == 0 {
                f[i][j] = f[i-1][j] + f[i][j-1]
            }
        }
    }
    return f[m-1][n-1]
}

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function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    const f = Array.from({ length: m }, () => Array(n).fill(0));
    for (let i = 0; i < m; i++) {
        if (obstacleGrid[i][0] === 1) {
            break;
        }
        f[i][0] = 1;
    }
    for (let i = 0; i < n; i++) {
        if (obstacleGrid[0][i] === 1) {
            break;
        }
        f[0][i] = 1;
    }
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            if (obstacleGrid[i][j] === 1) {
                continue;
            }
            f[i][j] = f[i - 1][j] + f[i][j - 1];
        }
    }
    return f[m - 1][n - 1];
}

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function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
    const m = obstacleGrid.length;
    const n = obstacleGrid[0].length;
    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
            return 0;
        }
        if (i === m - 1 && j === n - 1) {
            return 1;
        }
        if (f[i][j] === -1) {
            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
        }
        return f[i][j];
    };
    return dfs(0, 0);
}

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impl Solution {
    pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
        let m = obstacle_grid.len();
        let n = obstacle_grid[0].len();
        let mut f = vec![vec![0; n]; m];
        for i in 0..n {
            if obstacle_grid[0][i] == 1 {
                break;
            }
            f[0][i] = 1;
        }
        for i in 0..m {
            if obstacle_grid[i][0] == 1 {
                break;
            }
            f[i][0] = 1;
        }
        for i in 1..m {
            for j in 1..n {
                if obstacle_grid[i][j] == 1 {
                    continue;
                }
                f[i][j] = f[i - 1][j] + f[i][j - 1];
            }
        }
        f[m - 1][n - 1]
    }
}

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