Leetcode Unique Path II

最新推荐文章于 2026-01-05 22:36:07 发布

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本文探讨了在网格中寻找从左上角到右下角的唯一路径数量问题，并在此基础上增加了障碍物元素。通过动态规划的方法，给出了含障碍物情况下唯一路径数量的计算方案。

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.

跟之前的Unique Path的解法类似，动态规划大法。存在一点点区别，当obstacleGrid[i][j]为1时，dp[i][j] = 0；当obstacleGrid[i][j]为0时，其状态转移方程与前一题一模一样。

代码如下：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.empty())
            return 0;
        
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        
        int dp[m][n];
        memset(dp,0,sizeof(int)*m*n);
        dp[0][0] = 1;
        

    
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(obstacleGrid[i][j] == 1)
                {
                    dp[i][j] = 0;
                }
                else
                {
                    if(i-1 >= 0)
                        dp[i][j] += dp[i-1][j];
                    if(j-1 >= 0)
                        dp[i][j] += dp[i][j-1];
                }
            }
        }
        return dp[m-1][n-1];
    }
};