63. Unique Paths II

最新推荐文章于 2025-09-10 21:59:39 发布

转载最新推荐文章于 2025-09-10 21:59:39 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/9824784.html

文章标签：

#runtime #c/c++

本文探讨了在一个m x n网格中，机器人从左上角到右下角的路径规划问题，特别考虑了障碍物的存在。通过动态规划方法，计算在避开障碍物的情况下，机器人到达目标点的唯一路径数。

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [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

AC code:

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

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Unique Paths II.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9824784.html