LeetCode——Unique Paths II

最新推荐文章于 2025-09-11 19:19:55 发布

原创最新推荐文章于 2025-09-11 19:19:55 发布 · 392 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

Online Judge 专栏收录该内容

134 篇文章

订阅专栏

本文介绍了一个基于UniquePaths问题的变种——UniquePathsWithObstacles。该问题考虑了网格中存在障碍物的情况，探讨如何计算从起点到终点的不同路径数量。通过动态规划的方法，对存在障碍的网格进行特殊处理，确保算法正确计算出所有可行路径。

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.

» Solve this problem

在Unique Paths的基础上稍作修改即可，在grid上为1的点，其f值为0，表示无论如何都没法到达。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) {
            return 0;
        }
        
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        int f[m][n];
        memset(f, 0, sizeof(int) * m * n);
        f[0][0] = 1;
        
        for (int i = 1; i < m; i++) {
            f[i][0] = obstacleGrid[i][0] == 1 ? 0 : f[i - 1][0];
        }
        for (int i = 1; i < n; i++) {
            f[0][i] = obstacleGrid[0][i] == 1 ? 0 : f[0][i - 1];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    f[i][j] = 0;
                }
                else {
                    f[i][j] += obstacleGrid[i - 1][j] == 1 ? 0 : f[i - 1][j];
                    f[i][j] += obstacleGrid[i][j - 1] == 1 ? 0 : f[i][j - 1];                    
                }
            }
        }
        
        return f[m - 1][n - 1];
    }
};