diff --git a/lcci/16.04.Tic-Tac-Toe/README.md b/lcci/16.04.Tic-Tac-Toe/README.md
index bd3c28614bd6b..e073f53647620 100644
--- a/lcci/16.04.Tic-Tac-Toe/README.md
+++ b/lcci/16.04.Tic-Tac-Toe/README.md
@@ -5,6 +5,7 @@
## 题目描述
+
设计一个算法，判断玩家是否赢了井字游戏。输入是一个 N x N 的数组棋盘，由字符" "，"X"和"O"组成，其中字符" "代表一个空位。
以下是井字游戏的规则：
@@ -39,6 +40,17 @@
## 解法
+
+**方法一：计数**
+
+对于每个格子，如果是 `X`，我们不妨将计数加 $1$，如果是 `O`，我们不妨将计数减 $1$。那么当某个格子所在的行、列或者对角线的计数的绝对值等于 $n$ 时，说明当前玩家在该行、列或者对角线上放置了 $n$ 个相同字符，游戏结束，返回对应的字符即可。
+
+具体地，我们用一个长度为 $n$ 的一维数组 $rows$ 和 $cols$ 分别表示每一行和每一列的计数，用 $dg$ 和 $udg$ 分别表示两个对角线的计数。当玩家放置一个字符到 $(i, j)$ 时，根据字符是 `X` 还是 `O`，更新数组 $rows$，$cols$，$dg$ 以及 $udg$ 中对应元素的值。每次更新后，我们判断对应的元素的值的绝对值是否等于 $n$，如果等于 $n$，则说明当前玩家在行、列或者对角线上放置了 $n$ 个相同字符，游戏结束，返回对应的字符即可。
+
+最后，我们遍历整个棋盘，如果棋盘中存在字符 ` `，说明游戏还未结束，返回 `Pending`，否则返回 `Draw`。
+
+时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是棋盘的边长。
+
### **Python3**
@@ -46,7 +58,28 @@
```python
-
+class Solution:
+ def tictactoe(self, board: List[str]) -> str:
+ n = len(board)
+ rows = [0] * n
+ cols = [0] * n
+ dg = udg = 0
+ has_empty_grid = False
+ for i, row in enumerate(board):
+ for j, c in enumerate(row):
+ v = 1 if c == 'X' else -1
+ if c == ' ':
+ has_empty_grid = True
+ v = 0
+ rows[i] += v
+ cols[j] += v
+ if i == j:
+ dg += v
+ if i + j + 1 == n:
+ udg += v
+ if abs(rows[i]) == n or abs(cols[j]) == n or abs(dg) == n or abs(udg) == n:
+ return c
+ return 'Pending' if has_empty_grid else 'Draw'
```
### **Java**
@@ -54,7 +87,159 @@
```java
+class Solution {
+ public String tictactoe(String[] board) {
+ int n = board.length;
+ int[] rows = new int[n];
+ int[] cols = new int[n];
+ int dg = 0, udg = 0;
+ boolean hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i].charAt(j);
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (Math.abs(rows[i]) == n || Math.abs(cols[j]) == n || Math.abs(dg) == n
+ || Math.abs(udg) == n) {
+ return String.valueOf(c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+ string tictactoe(vector& board) {
+ int n = board.size();
+ vector rows(n), cols(n);
+ int dg = 0, udg = 0;
+ bool hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i][j];
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n) {
+ return string(1, c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+};
+```
+
+### **Go**
+
+```go
+func tictactoe(board []string) string {
+ n := len(board)
+ rows := make([]int, n)
+ cols := make([]int, n)
+ dg, udg := 0, 0
+ hasEmptyGrid := false
+ for i, row := range board {
+ for j, c := range row {
+ if c == ' ' {
+ hasEmptyGrid = true
+ continue
+ }
+ v := 1
+ if c == 'O' {
+ v = -1
+ }
+ rows[i] += v
+ cols[j] += v
+ if i == j {
+ dg += v
+ }
+ if i+j == n-1 {
+ udg += v
+ }
+ if abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n {
+ return string(c)
+ }
+ }
+ }
+ if hasEmptyGrid {
+ return "Pending"
+ }
+ return "Draw"
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
+```
+### **TypeScript**
+
+```ts
+function tictactoe(board: string[]): string {
+ const n = board.length;
+ const rows = Array(n).fill(0);
+ const cols = Array(n).fill(0);
+ let [dg, udg] = [0, 0];
+ let hasEmptyGrid = false;
+ for (let i = 0; i < n; ++i) {
+ for (let j = 0; j < n; ++j) {
+ const c = board[i][j];
+ if (c === ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ const v = c === 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i === j) {
+ dg += v;
+ }
+ if (i + j === n - 1) {
+ udg += v;
+ }
+ if (
+ Math.abs(rows[i]) === n ||
+ Math.abs(cols[j]) === n ||
+ Math.abs(dg) === n ||
+ Math.abs(udg) === n
+ ) {
+ return c;
+ }
+ }
+ }
+ return hasEmptyGrid ? 'Pending' : 'Draw';
+}
```
### **...**
diff --git a/lcci/16.04.Tic-Tac-Toe/README_EN.md b/lcci/16.04.Tic-Tac-Toe/README_EN.md
index 36fe3270d0b8e..6cddbe56c249a 100644
--- a/lcci/16.04.Tic-Tac-Toe/README_EN.md
+++ b/lcci/16.04.Tic-Tac-Toe/README_EN.md
@@ -51,18 +51,201 @@
## Solutions
+**Solution 1: Counting**
+
+For each cell, if it is `X`, we can add $1$ to the count; if it is `O`, we can subtract $1$ from the count. When the absolute value of the count of a row, column, or diagonal equals $n$, it means that the current player has placed $n$ identical characters in that row, column, or diagonal, and the game is over. We can return the corresponding character.
+
+Specifically, we use a one-dimensional array $rows$ and $cols$ of length $n$ to represent the count of each row and column, and use $dg$ and $udg$ to represent the count of the two diagonals. When a player places a character at $(i, j)$, we update the corresponding elements in the arrays $rows$, $cols$, $dg$, and $udg$ based on whether the character is `X` or `O`. After each update, we check whether the absolute value of the corresponding element equals $n$. If it does, it means that the current player has placed $n$ identical characters in that row, column, or diagonal, and the game is over. We can return the corresponding character.
+
+Finally, we traverse the entire board. If there is a character ` `, it means that the game is not over yet, and we return `Pending`. Otherwise, we return `Draw`.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the side length of the board.
+
### **Python3**
```python
-
+class Solution:
+ def tictactoe(self, board: List[str]) -> str:
+ n = len(board)
+ rows = [0] * n
+ cols = [0] * n
+ dg = udg = 0
+ has_empty_grid = False
+ for i, row in enumerate(board):
+ for j, c in enumerate(row):
+ v = 1 if c == 'X' else -1
+ if c == ' ':
+ has_empty_grid = True
+ v = 0
+ rows[i] += v
+ cols[j] += v
+ if i == j:
+ dg += v
+ if i + j + 1 == n:
+ udg += v
+ if abs(rows[i]) == n or abs(cols[j]) == n or abs(dg) == n or abs(udg) == n:
+ return c
+ return 'Pending' if has_empty_grid else 'Draw'
```
### **Java**
```java
+class Solution {
+ public String tictactoe(String[] board) {
+ int n = board.length;
+ int[] rows = new int[n];
+ int[] cols = new int[n];
+ int dg = 0, udg = 0;
+ boolean hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i].charAt(j);
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (Math.abs(rows[i]) == n || Math.abs(cols[j]) == n || Math.abs(dg) == n
+ || Math.abs(udg) == n) {
+ return String.valueOf(c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+ string tictactoe(vector& board) {
+ int n = board.size();
+ vector rows(n), cols(n);
+ int dg = 0, udg = 0;
+ bool hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i][j];
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n) {
+ return string(1, c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+};
+```
+
+### **Go**
+
+```go
+func tictactoe(board []string) string {
+ n := len(board)
+ rows := make([]int, n)
+ cols := make([]int, n)
+ dg, udg := 0, 0
+ hasEmptyGrid := false
+ for i, row := range board {
+ for j, c := range row {
+ if c == ' ' {
+ hasEmptyGrid = true
+ continue
+ }
+ v := 1
+ if c == 'O' {
+ v = -1
+ }
+ rows[i] += v
+ cols[j] += v
+ if i == j {
+ dg += v
+ }
+ if i+j == n-1 {
+ udg += v
+ }
+ if abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n {
+ return string(c)
+ }
+ }
+ }
+ if hasEmptyGrid {
+ return "Pending"
+ }
+ return "Draw"
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
+```
+
+### **TypeScript**
+```ts
+function tictactoe(board: string[]): string {
+ const n = board.length;
+ const rows = Array(n).fill(0);
+ const cols = Array(n).fill(0);
+ let [dg, udg] = [0, 0];
+ let hasEmptyGrid = false;
+ for (let i = 0; i < n; ++i) {
+ for (let j = 0; j < n; ++j) {
+ const c = board[i][j];
+ if (c === ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ const v = c === 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i === j) {
+ dg += v;
+ }
+ if (i + j === n - 1) {
+ udg += v;
+ }
+ if (
+ Math.abs(rows[i]) === n ||
+ Math.abs(cols[j]) === n ||
+ Math.abs(dg) === n ||
+ Math.abs(udg) === n
+ ) {
+ return c;
+ }
+ }
+ }
+ return hasEmptyGrid ? 'Pending' : 'Draw';
+}
```
### **...**
diff --git a/lcci/16.04.Tic-Tac-Toe/Solution.cpp b/lcci/16.04.Tic-Tac-Toe/Solution.cpp
new file mode 100644
index 0000000000000..f165fc4fe77b6
--- /dev/null
+++ b/lcci/16.04.Tic-Tac-Toe/Solution.cpp
@@ -0,0 +1,31 @@
+class Solution {
+public:
+ string tictactoe(vector& board) {
+ int n = board.size();
+ vector rows(n), cols(n);
+ int dg = 0, udg = 0;
+ bool hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i][j];
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n) {
+ return string(1, c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+};
\ No newline at end of file
diff --git a/lcci/16.04.Tic-Tac-Toe/Solution.go b/lcci/16.04.Tic-Tac-Toe/Solution.go
new file mode 100644
index 0000000000000..402e56ae2cc77
--- /dev/null
+++ b/lcci/16.04.Tic-Tac-Toe/Solution.go
@@ -0,0 +1,41 @@
+func tictactoe(board []string) string {
+ n := len(board)
+ rows := make([]int, n)
+ cols := make([]int, n)
+ dg, udg := 0, 0
+ hasEmptyGrid := false
+ for i, row := range board {
+ for j, c := range row {
+ if c == ' ' {
+ hasEmptyGrid = true
+ continue
+ }
+ v := 1
+ if c == 'O' {
+ v = -1
+ }
+ rows[i] += v
+ cols[j] += v
+ if i == j {
+ dg += v
+ }
+ if i+j == n-1 {
+ udg += v
+ }
+ if abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n {
+ return string(c)
+ }
+ }
+ }
+ if hasEmptyGrid {
+ return "Pending"
+ }
+ return "Draw"
+}
+
+func abs(x int) int {
+ if x < 0 {
+ return -x
+ }
+ return x
+}
\ No newline at end of file
diff --git a/lcci/16.04.Tic-Tac-Toe/Solution.java b/lcci/16.04.Tic-Tac-Toe/Solution.java
new file mode 100644
index 0000000000000..5ff5a378eff05
--- /dev/null
+++ b/lcci/16.04.Tic-Tac-Toe/Solution.java
@@ -0,0 +1,32 @@
+class Solution {
+ public String tictactoe(String[] board) {
+ int n = board.length;
+ int[] rows = new int[n];
+ int[] cols = new int[n];
+ int dg = 0, udg = 0;
+ boolean hasEmptyGrid = false;
+ for (int i = 0; i < n; ++i) {
+ for (int j = 0; j < n; ++j) {
+ char c = board[i].charAt(j);
+ if (c == ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ int v = c == 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i == j) {
+ dg += v;
+ }
+ if (i + j + 1 == n) {
+ udg += v;
+ }
+ if (Math.abs(rows[i]) == n || Math.abs(cols[j]) == n || Math.abs(dg) == n
+ || Math.abs(udg) == n) {
+ return String.valueOf(c);
+ }
+ }
+ }
+ return hasEmptyGrid ? "Pending" : "Draw";
+ }
+}
\ No newline at end of file
diff --git a/lcci/16.04.Tic-Tac-Toe/Solution.py b/lcci/16.04.Tic-Tac-Toe/Solution.py
new file mode 100644
index 0000000000000..c4c47432aac3d
--- /dev/null
+++ b/lcci/16.04.Tic-Tac-Toe/Solution.py
@@ -0,0 +1,27 @@
+class Solution:
+ def tictactoe(self, board: List[str]) -> str:
+ n = len(board)
+ rows = [0] * n
+ cols = [0] * n
+ dg = udg = 0
+ has_empty_grid = False
+ for i, row in enumerate(board):
+ for j, c in enumerate(row):
+ v = 1 if c == 'X' else -1
+ if c == ' ':
+ has_empty_grid = True
+ v = 0
+ rows[i] += v
+ cols[j] += v
+ if i == j:
+ dg += v
+ if i + j + 1 == n:
+ udg += v
+ if (
+ abs(rows[i]) == n
+ or abs(cols[j]) == n
+ or abs(dg) == n
+ or abs(udg) == n
+ ):
+ return c
+ return 'Pending' if has_empty_grid else 'Draw'
diff --git a/lcci/16.04.Tic-Tac-Toe/Solution.ts b/lcci/16.04.Tic-Tac-Toe/Solution.ts
new file mode 100644
index 0000000000000..fc04dde213cb2
--- /dev/null
+++ b/lcci/16.04.Tic-Tac-Toe/Solution.ts
@@ -0,0 +1,34 @@
+function tictactoe(board: string[]): string {
+ const n = board.length;
+ const rows = Array(n).fill(0);
+ const cols = Array(n).fill(0);
+ let [dg, udg] = [0, 0];
+ let hasEmptyGrid = false;
+ for (let i = 0; i < n; ++i) {
+ for (let j = 0; j < n; ++j) {
+ const c = board[i][j];
+ if (c === ' ') {
+ hasEmptyGrid = true;
+ continue;
+ }
+ const v = c === 'X' ? 1 : -1;
+ rows[i] += v;
+ cols[j] += v;
+ if (i === j) {
+ dg += v;
+ }
+ if (i + j === n - 1) {
+ udg += v;
+ }
+ if (
+ Math.abs(rows[i]) === n ||
+ Math.abs(cols[j]) === n ||
+ Math.abs(dg) === n ||
+ Math.abs(udg) === n
+ ) {
+ return c;
+ }
+ }
+ }
+ return hasEmptyGrid ? 'Pending' : 'Draw';
+}