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Commit 10e1921

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feat: add solutions to lcci problem: No.16.04 (#1760)
No.16.04.Tic-Tac-Toe
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‎lcci/16.04.Tic-Tac-Toe/README.md‎

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## 题目描述
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<!-- 这里写题目描述 -->
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<p>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&quot; &quot;,&quot;X&quot;和&quot;O&quot;组成,其中字符&quot; &quot;代表一个空位。</p>
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<p>以下是井字游戏的规则:</p>
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<ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:计数**
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对于每个格子,如果是 `X`,我们不妨将计数加 1ドル,ドル如果是 `O`,我们不妨将计数减 1ドル$。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。
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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 $rows,ドル$cols,ドル$dg$ 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,ドル如果等于 $n,ドル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。
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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`。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 是棋盘的边长。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def tictactoe(self, board: List[str]) -> str:
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n = len(board)
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rows = [0] * n
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cols = [0] * n
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dg = udg = 0
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has_empty_grid = False
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for i, row in enumerate(board):
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for j, c in enumerate(row):
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v = 1 if c == 'X' else -1
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if c == ' ':
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has_empty_grid = True
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v = 0
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rows[i] += v
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cols[j] += v
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if i == j:
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dg += v
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if i + j + 1 == n:
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udg += v
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if abs(rows[i]) == n or abs(cols[j]) == n or abs(dg) == n or abs(udg) == n:
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return c
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return 'Pending' if has_empty_grid else 'Draw'
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String tictactoe(String[] board) {
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int n = board.length;
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int[] rows = new int[n];
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int[] cols = new int[n];
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int dg = 0, udg = 0;
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boolean hasEmptyGrid = false;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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char c = board[i].charAt(j);
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if (c == ' ') {
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hasEmptyGrid = true;
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continue;
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}
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int v = c == 'X' ? 1 : -1;
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rows[i] += v;
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cols[j] += v;
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if (i == j) {
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dg += v;
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}
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if (i + j + 1 == n) {
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udg += v;
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}
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if (Math.abs(rows[i]) == n || Math.abs(cols[j]) == n || Math.abs(dg) == n
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|| Math.abs(udg) == n) {
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return String.valueOf(c);
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}
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}
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}
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return hasEmptyGrid ? "Pending" : "Draw";
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string tictactoe(vector<string>& board) {
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int n = board.size();
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vector<int> rows(n), cols(n);
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int dg = 0, udg = 0;
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bool hasEmptyGrid = false;
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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char c = board[i][j];
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if (c == ' ') {
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hasEmptyGrid = true;
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continue;
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}
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int v = c == 'X' ? 1 : -1;
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rows[i] += v;
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cols[j] += v;
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if (i == j) {
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dg += v;
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}
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if (i + j + 1 == n) {
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udg += v;
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}
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if (abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n) {
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return string(1, c);
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}
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}
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}
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return hasEmptyGrid ? "Pending" : "Draw";
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}
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};
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```
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### **Go**
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```go
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func tictactoe(board []string) string {
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n := len(board)
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rows := make([]int, n)
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cols := make([]int, n)
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dg, udg := 0, 0
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hasEmptyGrid := false
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for i, row := range board {
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for j, c := range row {
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if c == ' ' {
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hasEmptyGrid = true
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continue
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}
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v := 1
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if c == 'O' {
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v = -1
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}
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rows[i] += v
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cols[j] += v
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if i == j {
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dg += v
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}
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if i+j == n-1 {
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udg += v
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}
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if abs(rows[i]) == n || abs(cols[j]) == n || abs(dg) == n || abs(udg) == n {
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return string(c)
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}
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}
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}
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if hasEmptyGrid {
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return "Pending"
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}
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return "Draw"
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}
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func abs(x int) int {
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if x < 0 {
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return -x
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}
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return x
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}
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```
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### **TypeScript**
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```ts
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function tictactoe(board: string[]): string {
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const n = board.length;
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const rows = Array(n).fill(0);
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const cols = Array(n).fill(0);
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let [dg, udg] = [0, 0];
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let hasEmptyGrid = false;
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for (let i = 0; i < n; ++i) {
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for (let j = 0; j < n; ++j) {
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const c = board[i][j];
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if (c === ' ') {
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hasEmptyGrid = true;
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continue;
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}
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const v = c === 'X' ? 1 : -1;
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rows[i] += v;
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cols[j] += v;
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if (i === j) {
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dg += v;
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}
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if (i + j === n - 1) {
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udg += v;
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}
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if (
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Math.abs(rows[i]) === n ||
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Math.abs(cols[j]) === n ||
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Math.abs(dg) === n ||
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Math.abs(udg) === n
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) {
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return c;
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}
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}
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}
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return hasEmptyGrid ? 'Pending' : 'Draw';
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}
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```
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### **...**

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