【LeetCode】063. Unique Paths II-优快云博客

本文探讨了在一个含有障碍物的网格中计算从左上角到右下角的唯一路径数量的问题。通过动态规划的方法，给出了两种解决方案，并详细解释了如何在遇到障碍物时调整路径计数。

题目：

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.

题解：

Solution 1 ()(未优化）

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

Solution 2 ()

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