題目

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.

解題之法

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) return 0;
        vector<vector<int> > dp(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(), 0)); //注意初始化方法
        for (int i = 0; i < obstacleGrid.size(); ++i) {
            for (int j = 0; j < obstacleGrid[i].size(); ++j) {
                if (obstacleGrid[i][j] == 1) dp[i][j] = 0;
                else if (i == 0 && j == 0) dp[i][j] = 1;
                else if (i == 0 && j > 0) dp[i][j] = dp[i][j - 1];
                else if (i > 0 && j == 0) dp[i][j] = dp[i - 1][j];
                else dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp.back().back();
    }
};

分析

這道題是之前那道 Unique Paths 不同的路徑 的延伸，在路徑中加了一些障礙物，還是用動態規劃Dynamic Programming來解，不同的是當遇到為1的點，將該位置的dp數組中的值清零，其余和之前那道題并沒有什么區別。