#Leetcode# 63. Unique Paths II

最新推荐文章于 2025-08-22 01:16:35 发布

weixin_30517001

最新推荐文章于 2025-08-22 01:16:35 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/10003481.html

本文探讨了LeetCode上独特路径II问题的算法解决方案，通过动态规划方法，在存在障碍物的网格中寻找从起点到达终点的所有可能路径数量。特别关注了如何处理障碍物对路径的影响，提供了一种高效的计算方法。

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

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

代码：

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

$m$ 行 $n$ 列真的是写不习惯这个只多了一个是 $1$ 的时候走不动那就到该点的时候 $dp$ 设成 $0$ 就好了本来想用深搜搜路径数量但是。。。很难受写错了。。。晚上回去再好好撸 $dfs$ 吧

转载于:https://www.cnblogs.com/zlrrrr/p/10003481.html