Question

63. 不同路径 II

English Version

题目描述

一个机器人位于一个

 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish”）。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？

网格中的障碍物和空位置分别用 1 和 0 来表示。

 

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：动态规划

我们定义 $dp[i][j]$ 表示到达网格 $(i,j)$ 的路径数。

首先初始化 $dp$ 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 $dp[i][j] = 0$；
• 若 ￥ obstacleGrid[i][j] = 0$，则 $dp[i][j] = dp[i - 1][j] + dp[i][j - 1]$。

最后返回 $dp[m - 1][n - 1]$ 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。

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

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

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

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

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

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();
    if obstacle_grid[0][0] == 1 || obstacle_grid[m - 1][n - 1] == 1 {
      return 0;
    }
    let mut dp = vec![vec![0; n]; m];
    for i in 0..n {
      if obstacle_grid[0][i] == 1 {
        break;
      }
      dp[0][i] = 1;
    }
    for i in 0..m {
      if obstacle_grid[i][0] == 1 {
        break;
      }
      dp[i][0] = 1;
    }
    for i in 1..m {
      for j in 1..n {
        if obstacle_grid[i][j] == 1 {
          continue;
        }
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
      }
    }
    dp[m - 1][n - 1]
  }
}