2021-11-09：设计井字棋。谁先同行或者同列都是自己的棋子，就算获得胜利。力扣348。

答案2021-11-09：

记录n行，n列，对角线的某个棋手的棋子数量。

代码用golang编写。代码如下：

package main

import "fmt"

func main() {
    nttt := NewTicTacToe(3)
    fmt.Println(nttt.move(0, 0, 1))
    fmt.Println(nttt.move(1, 0, 1))
    fmt.Println(nttt.move(2, 0, 1))
}

type TicTacToe struct {
    rows    [][]int
    cols    [][]int
    leftUp  []int
    rightUp []int
    matrix  [][]bool
    N       int
}

func NewTicTacToe(n int) *TicTacToe {
    res := &TicTacToe{}
    // rows[a][1] : 1这个人，在a行上，下了几个
    // rows[b][2] : 2这个人，在b行上，下了几个
    //rows = new int[n][3]; //0 1 2
    res.rows = make([][]int, n)
    for i := 0; i < n; i++ {
        res.rows[i] = make([]int, 3)
    }

    res.cols = make([][]int, n)
    for i := 0; i < n; i++ {
        res.cols[i] = make([]int, 3)
    }
    // leftUp[2] = 7 : 2这个人，在左对角线上，下了7个
    res.leftUp = make([]int, 3)
    // rightUp[1] = 9 : 1这个人，在右对角线上，下了9个
    res.rightUp = make([]int, 3)
    res.matrix = make([][]bool, n)
    for i := 0; i < n; i++ {
        res.matrix[i] = make([]bool, n)
    }
    res.N = n
    return res
}

func (this *TicTacToe) move(row, col, player int) int {
    if this.matrix[row][col] {
        return 0
    }
    this.matrix[row][col] = true
    this.rows[row][player]++
    this.cols[col][player]++
    if row == col {
        this.leftUp[player]++
    }
    if row+col == this.N-1 {
        this.rightUp[player]++
    }
    if this.rows[row][player] == this.N || this.cols[col][player] == this.N || this.leftUp[player] == this.N || this.rightUp[player] == this.N {
        return player
    }
    return 0
}