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feat: add solutions to lc problems: No.1706,1707 #2132

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Dec 20, 2023
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feat: add solutions to lc problems: No.1706,1707 #2132

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754eb24
feat: add solutions to lc problem: No.1707
yanglbme Dec 20, 2023
142401e
feat: add solutions to lc problem: No.1706
yanglbme Dec 20, 2023
9e784d0
fix: cpp code
yanglbme Dec 20, 2023
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68 changes: 37 additions & 31 deletions solution/1700-1799/1706.Where Will the Ball Fall/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -66,14 +66,18 @@ b4 球开始放在第 4 列上，会卡在第 2、3 列和第 1 行之间的 "V"

<!-- 这里可写通用的实现逻辑 -->

球被卡住共有 4 种情况：
**方法一：分情况讨论 + DFS**

我们可以使用 DFS 来模拟球的运动过程，设计一个函数 $dfs(i, j)$，表示球从第 $i$ 行第 $j$ 列出发，最终会落在第几列。对于以下情况，球会卡住：

1. 球位于最左一列，并且球所在的单元格单元格挡板将球导向左侧
1. 球位于最右一列，并且此单元格挡板将球导向右侧
1. 球所在的单元格挡板将球导向右侧，并且球右侧相邻单元格挡板将球导向左侧
1. 球所在的单元格挡板将球导向左侧，并且球左侧相邻单元格挡板将球导向右侧

DFS 搜索即可。
如果满足以上任意一种情况，我们就可以判断球会卡住，返回 $-1$。否则，我们就可以继续递归地寻找球的下一个位置。最后，如果球到了最后一行，我们就可以返回当前列的编号。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是数组 $grid$ 的行数和列数。

<!-- tabs:start -->

Expand All @@ -84,10 +88,7 @@ DFS 搜索即可。
```python
class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])

def dfs(i, j):
nonlocal m, n
def dfs(i: int, j: int) -> int:
if i == m:
return j
if j == 0 and grid[i][j] == -1:
Expand All @@ -100,6 +101,7 @@ class Solution:
return -1
return dfs(i + 1, j + 1) if grid[i][j] == 1 else dfs(i + 1, j - 1)

m, n = len(grid), len(grid[0])
return [dfs(0, j) for j in range(n)]
```

Expand Down Expand Up @@ -150,35 +152,41 @@ class Solution {
```cpp
class Solution {
public:
int m, n;
vector<vector<int>> grid;

vector<int> findBall(vector<vector<int>>& grid) {
this->grid = grid;
m = grid.size();
n = grid[0].size();
int m = grid.size(), n = grid[0].size();
vector<int> ans(n);
for (int j = 0; j < n; ++j) ans[j] = dfs(0, j);
return ans;
}
function<int(int, int)> dfs = [&](int i, int j) {
if (i == m) {
return j;
}
if (j == 0 && grid[i][j] == -1) {
return -1;
}
if (j == n - 1 && grid[i][j] == 1) {
return -1;
}
if (grid[i][j] == 1 && grid[i][j + 1] == -1) {
return -1;
}
if (grid[i][j] == -1 && grid[i][j - 1] == 1) {
return -1;
}
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);

int dfs(int i, int j) {
if (i == m) return j;
if (j == 0 && grid[i][j] == -1) return -1;
if (j == n - 1 && grid[i][j] == 1) return -1;
if (grid[i][j] == 1 && grid[i][j + 1] == -1) return -1;
if (grid[i][j] == -1 && grid[i][j - 1] == 1) return -1;
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);
};
for (int j = 0; j < n; ++j) {
ans[j] = dfs(0, j);
}
return ans;
}
};
```

### **Go**

```go
func findBall(grid [][]int) []int {
func findBall(grid [][]int) (ans []int) {
m, n := len(grid), len(grid[0])

var dfs func(i, j int) int
dfs = func(i, j int) int {
if i == m {
Expand All @@ -201,12 +209,10 @@ func findBall(grid [][]int) []int {
}
return dfs(i+1, j-1)
}

var ans []int
for j := 0; j < n; j++ {
ans = append(ans, dfs(0, j))
}
return ans
return
}
```

Expand All @@ -216,7 +222,6 @@ func findBall(grid [][]int) []int {
function findBall(grid: number[][]): number[] {
const m = grid.length;
const n = grid[0].length;
const res = new Array(n).fill(0);
const dfs = (i: number, j: number) => {
if (i === m) {
return j;
Expand All @@ -233,10 +238,11 @@ function findBall(grid: number[][]): number[] {
return dfs(i + 1, j - 1);
}
};
for (let i = 0; i < n; i++) {
res[i] = dfs(0, i);
const ans: number[] = [];
for (let j = 0; j < n; ++j) {
ans.push(dfs(0, j));
}
return res;
return ans;
}
```

Expand Down
73 changes: 44 additions & 29 deletions solution/1700-1799/1706.Where Will the Ball Fall/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -60,17 +60,27 @@ Ball b4 is dropped at column 4 and will get stuck on the box between column 2 an

## Solutions

**Solution 1: Case Discussion + DFS**

We can use DFS to simulate the movement of the ball. We design a function $dfs(i, j)$, which represents that the ball starts from the $i$th row and the $j$th column, and finally falls in which column. The ball will get stuck in the following situations:

1. The ball is in the leftmost column, and the cell's vane directs the ball to the left.
2. The ball is in the rightmost column, and the cell's vane directs the ball to the right.
3. The cell's vane where the ball is located directs the ball to the right, and the vane of the cell adjacent to the right of the ball directs the ball to the left.
4. The cell's vane where the ball is located directs the ball to the left, and the vane of the cell adjacent to the left of the ball directs the ball to the right.

If any of the above situations are met, we can judge that the ball will get stuck and return $-1$. Otherwise, we can continue to recursively find the next position of the ball. Finally, if the ball reaches the last row, we can return the current column number.

The time complexity is $O(m \times n)$, and the space complexity is $O(m)$. Where $m$ and $n$ are the number of rows and columns of the array $grid$, respectively.

<!-- tabs:start -->

### **Python3**

```python
class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])

def dfs(i, j):
nonlocal m, n
def dfs(i: int, j: int) -> int:
if i == m:
return j
if j == 0 and grid[i][j] == -1:
Expand All @@ -83,6 +93,7 @@ class Solution:
return -1
return dfs(i + 1, j + 1) if grid[i][j] == 1 else dfs(i + 1, j - 1)

m, n = len(grid), len(grid[0])
return [dfs(0, j) for j in range(n)]
```

Expand Down Expand Up @@ -131,35 +142,41 @@ class Solution {
```cpp
class Solution {
public:
int m, n;
vector<vector<int>> grid;

vector<int> findBall(vector<vector<int>>& grid) {
this->grid = grid;
m = grid.size();
n = grid[0].size();
int m = grid.size(), n = grid[0].size();
vector<int> ans(n);
for (int j = 0; j < n; ++j) ans[j] = dfs(0, j);
return ans;
}
function<int(int, int)> dfs = [&](int i, int j) {
if (i == m) {
return j;
}
if (j == 0 && grid[i][j] == -1) {
return -1;
}
if (j == n - 1 && grid[i][j] == 1) {
return -1;
}
if (grid[i][j] == 1 && grid[i][j + 1] == -1) {
return -1;
}
if (grid[i][j] == -1 && grid[i][j - 1] == 1) {
return -1;
}
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);

int dfs(int i, int j) {
if (i == m) return j;
if (j == 0 && grid[i][j] == -1) return -1;
if (j == n - 1 && grid[i][j] == 1) return -1;
if (grid[i][j] == 1 && grid[i][j + 1] == -1) return -1;
if (grid[i][j] == -1 && grid[i][j - 1] == 1) return -1;
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);
};
for (int j = 0; j < n; ++j) {
ans[j] = dfs(0, j);
}
return ans;
}
};
```

### **Go**

```go
func findBall(grid [][]int) []int {
func findBall(grid [][]int) (ans []int) {
m, n := len(grid), len(grid[0])

var dfs func(i, j int) int
dfs = func(i, j int) int {
if i == m {
Expand All @@ -182,12 +199,10 @@ func findBall(grid [][]int) []int {
}
return dfs(i+1, j-1)
}

var ans []int
for j := 0; j < n; j++ {
ans = append(ans, dfs(0, j))
}
return ans
return
}
```

Expand All @@ -197,7 +212,6 @@ func findBall(grid [][]int) []int {
function findBall(grid: number[][]): number[] {
const m = grid.length;
const n = grid[0].length;
const res = new Array(n).fill(0);
const dfs = (i: number, j: number) => {
if (i === m) {
return j;
Expand All @@ -214,10 +228,11 @@ function findBall(grid: number[][]): number[] {
return dfs(i + 1, j - 1);
}
};
for (let i = 0; i < n; i++) {
res[i] = dfs(0, i);
const ans: number[] = [];
for (let j = 0; j < n; ++j) {
ans.push(dfs(0, j));
}
return res;
return ans;
}
```

Expand Down
51 changes: 29 additions & 22 deletions solution/1700-1799/1706.Where Will the Ball Fall/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,23 +1,30 @@
class Solution {
public:
int m, n;
vector<vector<int>> grid;

vector<int> findBall(vector<vector<int>>& grid) {
this->grid = grid;
m = grid.size();
n = grid[0].size();
vector<int> ans(n);
for (int j = 0; j < n; ++j) ans[j] = dfs(0, j);
return ans;
}

int dfs(int i, int j) {
if (i == m) return j;
if (j == 0 && grid[i][j] == -1) return -1;
if (j == n - 1 && grid[i][j] == 1) return -1;
if (grid[i][j] == 1 && grid[i][j + 1] == -1) return -1;
if (grid[i][j] == -1 && grid[i][j - 1] == 1) return -1;
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);
}
class Solution {
public:
vector<int> findBall(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> ans(n);
function<int(int, int)> dfs = [&](int i, int j) {
if (i == m) {
return j;
}
if (j == 0 && grid[i][j] == -1) {
return -1;
}
if (j == n - 1 && grid[i][j] == 1) {
return -1;
}
if (grid[i][j] == 1 && grid[i][j + 1] == -1) {
return -1;
}
if (grid[i][j] == -1 && grid[i][j - 1] == 1) {
return -1;
}
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);

};
for (int j = 0; j < n; ++j) {
ans[j] = dfs(0, j);
}
return ans;
}
};
7 changes: 2 additions & 5 deletions solution/1700-1799/1706.Where Will the Ball Fall/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,6 +1,5 @@
func findBall(grid [][]int) []int {
func findBall(grid [][]int) (ans []int) {
m, n := len(grid), len(grid[0])

var dfs func(i, j int) int
dfs = func(i, j int) int {
if i == m {
Expand All @@ -23,10 +22,8 @@ func findBall(grid [][]int) []int {
}
return dfs(i+1, j-1)
}

var ans []int
for j := 0; j < n; j++ {
ans = append(ans, dfs(0, j))
}
return ans
return
}
36 changes: 17 additions & 19 deletions solution/1700-1799/1706.Where Will the Ball Fall/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,17 @@
class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])

def dfs(i, j):
nonlocal m, n
if i == m:
return j
if j == 0 and grid[i][j] == -1:
return -1
if j == n - 1 and grid[i][j] == 1:
return -1
if grid[i][j] == 1 and grid[i][j + 1] == -1:
return -1
if grid[i][j] == -1 and grid[i][j - 1] == 1:
return -1
return dfs(i + 1, j + 1) if grid[i][j] == 1 else dfs(i + 1, j - 1)

return [dfs(0, j) for j in range(n)]
class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
def dfs(i: int, j: int) -> int:
if i == m:
return j
if j == 0 and grid[i][j] == -1:
return -1
if j == n - 1 and grid[i][j] == 1:
return -1
if grid[i][j] == 1 and grid[i][j + 1] == -1:
return -1
if grid[i][j] == -1 and grid[i][j - 1] == 1:
return -1
return dfs(i + 1, j + 1) if grid[i][j] == 1 else dfs(i + 1, j - 1)

m, n = len(grid), len(grid[0])
return [dfs(0, j) for j in range(n)]
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