[Leetcode] 63. Unique Paths II 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/54598186

本文探讨了在包含障碍物的网格中寻找从左上角到右下角的唯一路径数量的问题。通过动态规划的方法，文章提供了一种高效的解决方案，并详细介绍了实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Follow up for "Unique Paths":

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 3x3 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.

思路：

当obstacleGrid[i][j] == 1时，将dp[i][j]直接设置为0；否则递推关系和Leetcode 62完全相同。其动态规划解法的时间复杂度为O(m*n)，空间复杂度还可以优化到O(n)。如果再要继续较真，空间复杂度还可以进一步优化到O(min(m, n))。其中m和n分别是obstacleGrid的列个数和行个数。

代码：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int rows = obstacleGrid.size();
        if (rows == 0) {
            return 0;
        } 
        int cols = obstacleGrid[0].size();
        if (cols == 0) {
            return 0;
        }
        vector<int> dp(cols + 1, 0);
        dp[1] = 1;
        for (int y = 1; y <= rows; ++y) {
            for (int x = 1; x <= cols; ++x) {
                if (obstacleGrid[y - 1][x - 1] == 1) {
                    dp[x] = 0;
                }
                else {
                    dp[x] = dp[x - 1] + dp[x];
                }
            }
        }
        return dp[cols];
    }
};