LeetCode-Unique Paths II

最新推荐文章于 2025-09-09 19:16:33 发布

转载最新推荐文章于 2025-09-09 19:16:33 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/superzrx/p/3269187.html

文章标签：

#c/c++

本文介绍了一个基于动态规划的算法，用于解决在一个包含障碍物的网格中寻找从左上角到右下角的不同路径数量的问题。网格中的每个单元格要么为空（0），要么为障碍物（1）。文章给出了详细的C++实现代码。

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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<long long>> dp;
        
        for(unsigned int i=0;i<obstacleGrid.size();i++){
            vector<long long>row;
            for(unsigned int j=0;j<obstacleGrid[0].size();j++){
                row.push_back(0);
            }
            dp.push_back(row);
        }
        dp[0][0]=1;
         for(unsigned int i=0;i<obstacleGrid.size();i++){
            for(unsigned int j=0;j<obstacleGrid[0].size();j++){
                if(!obstacleGrid[i][j]){
                    if(i>0)dp[i][j]+=dp[i-1][j];
                    if(j>0)dp[i][j]+=dp[i][j-1];
                }
                else dp[i][j]=0;
            }
        }
        return dp[obstacleGrid.size()-1][obstacleGrid[0].size()-1];
    }
};

转载于:https://www.cnblogs.com/superzrx/p/3269187.html