Unique Paths II(不同的路徑總數之二)

【難度：Medium】
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.

解題思路

在62題的基礎上解決這個問題，但需要考慮的是，機器人不能簡單地向下或向右走，要判斷是否有障礙物，因此左邊界和上邊界的初始化也要根據上一個點的狀態來設定，同樣使用動態規劃來完成。

c++程式碼如下：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if (obstacleGrid.empty())
            return
 
 0;
        //入口有障礙的話，直接返回
        if (obstacleGrid[0][0] == 1)
            return 0;
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        vector<vector<int>> path(m,vector<int>(n,0));
        //入口點走法只有1種。
        path[0][0] = 1;
        for (int i = 1; i < m; i++) {
            //左邊界，如果上一點可行並且當前點沒有障礙物，那麼該點可走
 

            if (path[i-1][0] != 0 && obstacleGrid[i][0] != 1)
                path[i][0] = 1;

        }
        for (int i = 1; i < n; i++) {
            //上邊界與左邊界情況同理
            if (path[0][i-1] != 0 && obstacleGrid[0][i] != 1)
                path[0][i] = 1;
        }
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                //動態規劃，當前點無障礙物則與62題處理方法一致
                if (obstacleGrid[i][j] != 1)
                    path[i][j] = path[i-1][j]+path[i][j-1];
        return path[m-1][n-1];
    }
};