題目描述

999. 可以被一步捕獲的棋子數(shù) - 力扣（LeetCode）

在一個 8 x 8 的棋盤上，有一個白色的車（Rook），用字符 'R' 表示。棋盤上還可能存在空方塊，白色的象（Bishop）以及黑色的卒（pawn），分別用字符 '.'，'B' 和 'p' 表示。不難看出，大寫字符表示的是白棋，小寫字符表示的是黑棋。

車按國際象棋中的規(guī)則移動。東，西，南，北四個基本方向任選其一，然后一直向選定的方向移動，直到滿足下列四個條件之一：

• 棋手選擇主動停下來。
• 棋子因到達棋盤的邊緣而停下。
• 棋子移動到某一方格來捕獲位于該方格上敵方（黑色）的卒，停在該方格內(nèi)。
• 車不能進入/越過已經(jīng)放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你現(xiàn)在可以控制車移動一次，請你統(tǒng)計有多少敵方的卒處于你的捕獲范圍內(nèi)（即，可以被一步捕獲的棋子數(shù)）。

示例 1：

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋：
在本例中，車能夠捕獲所有的卒。

示例 2：

輸入：[[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：0
解釋：
象阻止了車捕獲任何卒。

示例 3：

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋： 
車可以捕獲位置 b5，d6 和 f5 的卒。

提示：

board.length == board[i].length == 8

board[i][j] 可以是 'R'，'.'，'B' 或 'p'

只有一個格子上存在 board[i][j] == 'R'

思路分析

這道題首先要理解題意

• 如果沒有阻擋，車可以無限移動，除非自己停止
• 遇到象，停止。停止的意思是不能向前，但可以向后
• 遇到邊，停止。停止的意思是不能向前，但可以向后
• 遇到卒，吃掉，然后在這個方向上必須停止。

解題方法，先整理

• 去掉沒用的信息
• 把二維問題轉為一維問題。

AC 代碼

/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function (board) {
  let count = 0
  let info = []
  for (let i = 0; i < 8; i++) {
    let item = []
    for (let j = 0; j < 8; j++) {
      if ('.' !== board[i][j]) {
        item.push(board[i][j])
      }
    }
    item.length > 0 && info.push(item)
  }
  for (let j = 0; j < 8; j++) {
    let item = []
    for (let i = 0; i < 8; i++) {
      if ('.' !== board[i][j]) {
        item.push(board[i][j])
      }
    }
    item.length > 0 && info.push(item)
  }
  //整理好后的info是個一維數(shù)組
  for (let item of info) {
    let index = item.indexOf('R')
    if (index < 0) continue
    let i = index
    while (i--) {
      if (item[i] === 'B') break;
      if (item[i] === 'p') {
        count++
        break
      }
    }
    i = index + 1
    while (i < item.length) {
      if (item[i] === 'B') break;
      if (item[i] === 'p') {
        count++
        break
      }
      i++
    }
  }
  return count
};

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