題1 車的可用捕獲量

描述

在一個 8 x 8 的棋盤上，有一個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字符 “R”，“.”，“B” 和 “p” 給出。大寫字符表示白棋，小寫字符表示黑棋。
車按國際象棋中的規(guī)則移動：它選擇四個基本方向中的一個（北，東，西和南），然后朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進入同一個方格。
返回車能夠在一次移動中捕獲到的卒的數(shù)量。
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。
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’

題解

思路：棋子可以朝前后左右走N步，走到邊緣或者遇到B和p停止。用dx和dy來控制走的方向，（-1，0）左（1，0）右（0，-1）上（0，1）下，x，y表示R的坐標(biāo)，先遍歷棋盤找到R的坐標(biāo)位置i，j。在while循環(huán)中行走N步。

class Solution {public int numRookCaptures(char[][] board) {int[] dx = {-1, 1, 0, 0};int[] dy = {0, 0, -1, 1};for(int i = 0; i < 8; i++){for(int j = 0; j < 8; j++){if(board[i][j] == 'R'){int ans = 0;for(int k = 0; k < 4; k++){int x = i, y = j;while(true){x += dx[k];y += dy[k];if(x < 0||y < 0||x >= 8||y >= 8||board[x][y] == 'B'){break;}if(board[x][y] == 'p'){ans++;break;}}}return ans;}}}return 0;} }

總結(jié)

以上是生活随笔為你收集整理的【LeetCode】3月26日打卡-Day11的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。