63. Unique Paths II

Problem

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 10⁹.

https://leetcode.cn/problems/unique-paths-ii/

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Constraints:

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

Test Cases

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:

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import pytest

from solution import Solution


@pytest.mark.parametrize('obstacleGrid, expected', [
    ([[0,0,0],[0,1,0],[0,0,0]], 2),
    ([[0,1],[0,0]], 1),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, obstacleGrid, expected):
    assert sol.uniquePathsWithObstacles(obstacleGrid) == expected

Thoughts

62. Unique Paths 的进阶版，有些位置会有障碍物。

因为障碍物是随意摆放的，就没法直接用组合数计算了，可以按照普通的动态规划的方法计算，根据障碍物的特性调整一下状态转移公式。

显然如果格子 (i, j) 有障碍物，那么 u[i][j] 就只能是 0，否则就按照 62. Unique Paths 中的递推式子计算即可。

时间复杂度 O(m * n)，空间复杂度 O(n)。

Code

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class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: list[list[int]]) -> int:
        m = len(obstacleGrid)
        n = len(obstacleGrid[0])
        u = [0] * n
        u[0] = 1

        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j] == 1:
                    u[j] = 0
                elif j > 0:
                    u[j] += u[j-1]

        return u[-1]

参考资料

• 本文引用

  • 62. Unique Paths

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

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