【LeetCode】Available Captures for Rook(车的可用捕获量)

这道题是LeetCode里的第999道题。

题目叙述：

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：

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

这道题很简单，首先我们要找到车的位置在哪，然后从车这个位置开始遍历它的上下左右方向的格子，判断是否为象或者卒。

代码如下：

class Solution {
    public int numRookCaptures(char[][] board) {
        		int R_x=-1,R_y=-1;
		int res=0;
		//find R
		for(int i=0;i<8;i++) {
			for(int j=0;j<8;j++) {
				if(board[i][j]=='R') {
					R_x=i;R_y=j;break;
				}
			}
			if(R_x>=0 && R_y>=0)break;
		}
		//up,down,left,right
		for(int i=1;i+R_y<8;i++) {
			if(board[R_x][i+R_y]=='B')break;
			else if(board[R_x][i+R_y]=='p'){res++;break;}
		}
		for(int i=1;R_y-i>=0;i++) {
			if(board[R_x][R_y-i]=='B')break;
			else if(board[R_x][R_y-i]=='p') {res++;break;}
		}
		for(int i=1;R_x+i<8;i++) {
			if(board[R_x+i][R_y]=='B')break;
			else if(board[R_x+i][R_y]=='p') {res++;break;}
		}
		for(int i=1;R_x-i>=0;i++) {
			if(board[R_x-i][R_y]=='B')break;
			else if(board[R_x-i][R_y]=='p'){res++;break;}
		}
		return res;
    }
}

提交结果：

个人总结：

这题数组题真的太简单了，被第1000题虐后来这里找找自信！