feat: add solutions to lc problem: No.63 (doocs#2749)

Author: yanglbme

solution/0000-0099/0063.Unique Paths II/README.md

@@ -49,16 +49,151 @@
 
 ## 解法
 
-### 方法一：动态规划
+### 方法一：记忆化搜索
 
-我们定义 $dp[i][j]$ 表示到达网格 $(i,j)$ 的路径数。
+我们设计一个函数 $dfs(i, j)$ 表示从网格 $(i, j)$ 到网格 $(m - 1, n - 1)$ 的路径数。其中 $m$ 和 $n$ 分别是网格的行数和列数。
 
-首先初始化 $dp$ 第一列和第一行的所有值，然后遍历其它行和列，有两种情况：
+函数 $dfs(i, j)$ 的执行过程如下：
 
--   若 $obstacleGrid[i][j] = 1$，说明路径数为 $0$，那么 $dp[i][j] = 0$；
--   若 ￥ obstacleGrid[i][j] = 0$，则 $dp[i][j] = dp[i - 1][j] + dp[i][j - 1]$。
+-   如果 $i \ge m$ 或者 $j \ge n$，或者 $obstacleGrid[i][j] = 1$，则路径数为 $0$；
+-   如果 $i = m - 1$ 且 $j = n - 1$，则路径数为 $1$；
+-   否则，路径数为 $dfs(i + 1, j) + dfs(i, j + 1)$。
 
-最后返回 $dp[m - 1][n - 1]$ 即可。
+为了避免重复计算，我们可以使用记忆化搜索的方法。
+
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。
+
+<!-- tabs:start -->
+
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        @cache
+        def dfs(i: int, j: int) -> int:
+            if i >= m or j >= n or obstacleGrid[i][j]:
+                return 0
+            if i == m - 1 and j == n - 1:
+                return 1
+            return dfs(i + 1, j) + dfs(i, j + 1)
+
+        m, n = len(obstacleGrid), len(obstacleGrid[0])
+        return dfs(0, 0)
+```
+
+```java
+class Solution {
+    private Integer[][] f;
+    private int[][] obstacleGrid;
+    private int m;
+    private int n;
+
+    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
+        m = obstacleGrid.length;
+        n = obstacleGrid[0].length;
+        this.obstacleGrid = obstacleGrid;
+        f = new Integer[m][n];
+        return dfs(0, 0);
+    }
+
+    private int dfs(int i, int j) {
+        if (i >= m || j >= n || obstacleGrid[i][j] == 1) {
+            return 0;
+        }
+        if (i == m - 1 && j == n - 1) {
+            return 1;
+        }
+        if (f[i][j] == null) {
+            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
+        }
+        return f[i][j];
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
+        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
+        int f[m][n];
+        memset(f, -1, sizeof(f));
+        function<int(int, int)> dfs = [&](int i, int j) {
+            if (i >= m || j >= n || obstacleGrid[i][j]) {
+                return 0;
+            }
+            if (i == m - 1 && j == n - 1) {
+                return 1;
+            }
+            if (f[i][j] == -1) {
+                f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
+            }
+            return f[i][j];
+        };
+        return dfs(0, 0);
+    }
+};
+```
+
+```go
+func uniquePathsWithObstacles(obstacleGrid [][]int) int {
+	m, n := len(obstacleGrid), len(obstacleGrid[0])
+	f := make([][]int, m)
+	for i := range f {
+		f[i] = make([]int, n)
+		for j := range f[i] {
+			f[i][j] = -1
+		}
+	}
+	var dfs func(i, j int) int
+	dfs = func(i, j int) int {
+		if i >= m || j >= n || obstacleGrid[i][j] == 1 {
+			return 0
+		}
+		if i == m-1 && j == n-1 {
+			return 1
+		}
+		if f[i][j] == -1 {
+			f[i][j] = dfs(i+1, j) + dfs(i, j+1)
+		}
+		return f[i][j]
+	}
+	return dfs(0, 0)
+}
+```
+
+```ts
+function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
+    const m = obstacleGrid.length;
+    const n = obstacleGrid[0].length;
+    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
+    const dfs = (i: number, j: number): number => {
+        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
+            return 0;
+        }
+        if (i === m - 1 && j === n - 1) {
+            return 1;
+        }
+        if (f[i][j] === -1) {
+            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
+        }
+        return f[i][j];
+    };
+    return dfs(0, 0);
+}
+```
+
+<!-- tabs:end -->
+
+### 方法二：动态规划
+
+我们定义 $f[i][j]$ 表示到达网格 $(i,j)$ 的路径数。
+
+首先初始化 $f$ 第一列和第一行的所有值，然后遍历其它行和列，有两种情况：
+
+-   若 $obstacleGrid[i][j] = 1$，说明路径数为 $0$，那么 $f[i][j] = 0$；
+-   若 $obstacleGrid[i][j] = 0$，则 $f[i][j] = f[i - 1][j] + f[i][j - 1]$。
+
+最后返回 $f[m - 1][n - 1]$ 即可。
 
 时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。
 
@@ -68,41 +203,41 @@
 class Solution:
     def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
         m, n = len(obstacleGrid), len(obstacleGrid[0])
-        dp = [[0] * n for _ in range(m)]
+        f = [[0] * n for _ in range(m)]
         for i in range(m):
             if obstacleGrid[i][0] == 1:
                 break
-            dp[i][0] = 1
+            f[i][0] = 1
         for j in range(n):
             if obstacleGrid[0][j] == 1:
                 break
-            dp[0][j] = 1
+            f[0][j] = 1
         for i in range(1, m):
             for j in range(1, n):
                 if obstacleGrid[i][j] == 0:
-                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
-        return dp[-1][-1]
+                    f[i][j] = f[i - 1][j] + f[i][j - 1]
+        return f[-1][-1]
 ```
 
 ```java
 class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
         int m = obstacleGrid.length, n = obstacleGrid[0].length;
-        int[][] dp = new int[m][n];
+        int[][] f = new int[m][n];
         for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
-            dp[i][0] = 1;
+            f[i][0] = 1;
         }
         for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
-            dp[0][j] = 1;
+            f[0][j] = 1;
         }
         for (int i = 1; i < m; ++i) {
             for (int j = 1; j < n; ++j) {
                 if (obstacleGrid[i][j] == 0) {
-                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+                    f[i][j] = f[i - 1][j] + f[i][j - 1];
                 }
             }
         }
-        return dp[m - 1][n - 1];
+        return f[m - 1][n - 1];
     }
 }
 ```
@@ -112,75 +247,96 @@ class Solution {
 public:
     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
         int m = obstacleGrid.size(), n = obstacleGrid[0].size();
-        vector<vector<int>> dp(m, vector<int>(n));
+        vector<vector<int>> f(m, vector<int>(n));
         for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
-            dp[i][0] = 1;
+            f[i][0] = 1;
         }
         for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
-            dp[0][j] = 1;
+            f[0][j] = 1;
         }
         for (int i = 1; i < m; ++i) {
             for (int j = 1; j < n; ++j) {
                 if (obstacleGrid[i][j] == 0) {
-                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+                    f[i][j] = f[i - 1][j] + f[i][j - 1];
                 }
             }
         }
-        return dp[m - 1][n - 1];
+        return f[m - 1][n - 1];
     }
 };
 ```
 
 ```go
 func uniquePathsWithObstacles(obstacleGrid [][]int) int {
 	m, n := len(obstacleGrid), len(obstacleGrid[0])
-	dp := make([][]int, m)
+	f := make([][]int, m)
 	for i := 0; i < m; i++ {
-		dp[i] = make([]int, n)
+		f[i] = make([]int, n)
 	}
 	for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
-		dp[i][0] = 1
+		f[i][0] = 1
 	}
 	for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {
-		dp[0][j] = 1
+		f[0][j] = 1
 	}
 	for i := 1; i < m; i++ {
 		for j := 1; j < n; j++ {
 			if obstacleGrid[i][j] == 0 {
-				dp[i][j] = dp[i-1][j] + dp[i][j-1]
+				f[i][j] = f[i-1][j] + f[i][j-1]
 			}
 		}
 	}
-	return dp[m-1][n-1]
+	return f[m-1][n-1]
 }
 ```
 
 ```ts
 function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
     const m = obstacleGrid.length;
     const n = obstacleGrid[0].length;
-    const dp = Array.from({ length: m }, () => new Array(n).fill(0));
+    const f = Array.from({ length: m }, () => Array(n).fill(0));
     for (let i = 0; i < m; i++) {
         if (obstacleGrid[i][0] === 1) {
             break;
         }
-        dp[i][0] = 1;
+        f[i][0] = 1;
     }
     for (let i = 0; i < n; i++) {
         if (obstacleGrid[0][i] === 1) {
             break;
         }
-        dp[0][i] = 1;
+        f[0][i] = 1;
     }
     for (let i = 1; i < m; i++) {
         for (let j = 1; j < n; j++) {
             if (obstacleGrid[i][j] === 1) {
                 continue;
             }
-            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+            f[i][j] = f[i - 1][j] + f[i][j - 1];
         }
     }
-    return dp[m - 1][n - 1];
+    return f[m - 1][n - 1];
+}
+```
+
+```ts
+function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
+    const m = obstacleGrid.length;
+    const n = obstacleGrid[0].length;
+    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(-1));
+    const dfs = (i: number, j: number): number => {
+        if (i >= m || j >= n || obstacleGrid[i][j] === 1) {
+            return 0;
+        }
+        if (i === m - 1 && j === n - 1) {
+            return 1;
+        }
+        if (f[i][j] === -1) {
+            f[i][j] = dfs(i + 1, j) + dfs(i, j + 1);
+        }
+        return f[i][j];
+    };
+    return dfs(0, 0);
 }
 ```
 
@@ -189,31 +345,28 @@ impl Solution {
     pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
         let m = obstacle_grid.len();
         let n = obstacle_grid[0].len();
-        if obstacle_grid[0][0] == 1 || obstacle_grid[m - 1][n - 1] == 1 {
-            return 0;
-        }
-        let mut dp = vec![vec![0; n]; m];
+        let mut f = vec![vec![0; n]; m];
         for i in 0..n {
             if obstacle_grid[0][i] == 1 {
                 break;
             }
-            dp[0][i] = 1;
+            f[0][i] = 1;
         }
         for i in 0..m {
             if obstacle_grid[i][0] == 1 {
                 break;
             }
-            dp[i][0] = 1;
+            f[i][0] = 1;
         }
         for i in 1..m {
             for j in 1..n {
                 if obstacle_grid[i][j] == 1 {
                     continue;
                 }
-                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+                f[i][j] = f[i - 1][j] + f[i][j - 1];
             }
         }
-        dp[m - 1][n - 1]
+        f[m - 1][n - 1]
     }
 }
 ```