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+## 何も見ずに解いてみる
+
+```cpp
+class Solution {
+public:
+    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
+        vector<vector<int>> num_paths(obstacleGrid.size(), vector<int>(obstacleGrid[0].size(), 0));
+        if (obstacleGrid[0][0] == 1) {
+            return 0;
+        }
+        num_paths[0][0] = 1;
+        for (int i = 0; i < obstacleGrid.size(); i++) {
+            for (int j = 0; j < obstacleGrid[0].size(); j++) {
+                if (obstacleGrid[i][j] == 1) {
+                    continue;
+                }
+                if (i != 0) {
+                    num_paths[i][j] += num_paths[i-1][j];
+                }
+                if (j != 0) {
+                    num_paths[i][j] += num_paths[i][j-1];
+                }
+            }
+        }
+        return num_paths.back().back();
+    }
+};
+```
+
+## 他の人の解答を見てみる
+
+https://github.com/ryosuketc/leetcode_arai60/pull/47/files  
+https://github.com/tokuhirat/LeetCode/pull/34/files  
+https://github.com/irohafternoon/LeetCode/pull/37/files  
+https://github.com/philip82148/leetcode-swejp/pull/10/files  
+
+最初の行と列を別の枠で初期化しているやり方をしているコードも結構あるが、個人的には`i != 0`や`i > 0`で条件分岐するやり方のほうが見やすいと感じた。
+
+列のサイズの配列で行うバージョン
+
+```cpp
+class Solution {
+public:
+    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
+        int num_rows = obstacleGrid.size();
+        int num_columns = obstacleGrid[0].size();
+        vector<int> num_paths_current_row(num_columns, 0);
+        num_paths_current_row[0] = 1;
+        for (int i = 0; i < num_rows; i++) {
+            for (int j = 0; j < num_columns; j++) {
+                if (obstacleGrid[i][j] == 1) {
+                    num_paths_current_row[j] = 0;
+                    continue;
+                }
+                if (j > 0) {
+                    num_paths_current_row[j] += num_paths_current_row[j - 1];
+                }
+            }
+        }
+        return num_paths_current_row.back();
+    }
+};
+```
+
+## 最終コード
+
+```cpp
+class Solution {
+public:
+    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
+        int num_rows = obstacleGrid.size();
+        int num_columns = obstacleGrid[0].size();
+        vector<vector<int>> num_paths(num_rows, vector<int>(num_columns, 0));
+        if (obstacleGrid[0][0] == 1) {
+            return 0;
+        }
+        num_paths[0][0] = 1;
+        for (int i = 0; i < num_rows; i++) {
+            for (int j = 0; j < num_columns; j++) {
+                if (obstacleGrid[i][j] == 1) {
+                    continue;
+                }
+                if (i != 0) {
+                    num_paths[i][j] += num_paths[i-1][j];
+                }
+                if (j != 0) {
+                    num_paths[i][j] += num_paths[i][j-1];
+                }
+            }
+        }
+        return num_paths.back().back();
+    }
+};
+```