每日两题 T6 - 码农教程

算法

LeetCode T999. 车的可用捕获量[1]

描述

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。

示例 1：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

示例 2：

输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

示例 3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
车可以捕获位置 b5，d6 和 f5 的卒。

提示：

•board.length == board[i].length == 8•board[i][j] 可以是 'R'，'.'，'B' 或 'p'•只有一个格子上存在 board[i][j] == 'R'

分析

没下过国际象棋，题目有读不懂，放弃！

其实，描述说的问题我们通过图来说明，以白色车为原点，向上下、左右试探。

所以我们首先找到白色车所在位置，然后遍历获取到 X 向、Y 向数组，并且过滤掉空白块，得到的数组我们只需要从第一个"p"开始找，到下一个"p"或者墙结束

代码

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function(board) {
    let y, xlist = [], ylist = [], boardXLen = 8;

    for (let i = 0; i < boardXLen; i++) {
      if (board[i].indexOf('R') > -1) {
        y = board[i].indexOf('R')
        xlist = board[i].filter(e => e !== '.')
      }
    }
    for (let i = 0; i < boardXLen; i++) {
      board[i][y] !== '.' && ylist.push(board[i][y])
    }

    let rx = xlist.indexOf('R'), ry = ylist.indexOf('R'), num = 0

    if(xlist[rx+1] && xlist[rx+1] === 'p') ++num
    if(xlist[rx-1] && xlist[rx-1] === 'p') ++num
    if(ylist[ry+1] && ylist[ry+1] === 'p') ++num
    if(ylist[ry-1] && ylist[ry-1] === 'p') ++num

    return num

};

JavaScript

ES5和ES6 的继承有什么区别？

•class 声明变量会提升，但不会初始化赋值。变量进入暂时性死区，类似于 let、const 声明

const p = new People(); // it's ok
function People() {
  this.bar = 1;
}

const m = new Man(); // ReferenceError: Foo is not defined
class Man {
  constructor() {
    this.foo = 1;
  }
}

•class 声明内部会启用严格模式

function People() {
  baz = 1; // it's ok
}
const p = new People();

class Man {
  constructor() {
    fol = 1; // ReferenceError: fol is not defined
  }
}
const m = new Man();

•class 的所有方法（包括静态方法和实例方法）是不可枚举

// 引用一个未声明的变量
function People() {
  this.bar = 1;
}
People.say = function() {
  return 1;
};
People.prototype.eat = function() {
    // ...
};
const pKeys = Object.keys(Bar); // ['say']
const pProtoKeys = Object.keys(Bar.prototype); // ['eat']

class Man {
  constructor() {
    this.foo = 1;
  }
  static say() {
    return 1;
  }
  eat() {
        // ...
  }
}
const mKeys = Object.keys(Man); // []
const mProtoKeys = Object.keys(Man.prototype); // []

•class 的所有方法（包括静态方法和实例方法）都没有原型对象 prototype，所以也没有[[construct]]，不能使用 new 来调用。

function People() {
  this.bar = 1;
}
People.prototype.print = function() {
  console.log(this.bar);
};

const p = new People();
const pPrint = new bar.print(); // it's ok

class Man {
  constructor() {
    this.foo = 42;
  }
  print() {
    console.log(this.foo);
  }
}
const m = new Man();
const mPrint = new m.print(); // TypeError: foo.print is not a constructor

•必须使用 new 调用 class。

function People() {
  this.bar = 1;
}
const p = People(); // it's ok

class Man {
  constructor() {
    this.foo = 1;
  }
}
const m = Man(); // TypeError: Class constructor Foo cannot be invoked without 'new'

•class 内部无法重写类名。

function People() {
  People = 'Pap'; // it's ok
  this.bar = 1;
}
const p = new People();
// People: 'Pap'
// bar: People {bar: 1}  

class Man {
  constructor() {
    this.foo = 42;
    Man = 'Woman'; // TypeError: Assignment to constant variable
  }
}
const m = new Man();
Man = 'Fol'; // it's ok

References

[1] 999. 车的可用捕获量: https://leetcode-cn.com/problems/available-captures-for-rook/