LeetCode 999. 车的可用捕获量

999. 车的可用捕获量

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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'

思路：刚开始首先想到的是深度搜索，后来发现用四个for就可以解决这个问题，首先遍历数组，找到车的位置，然后向四个方向进行搜索，就可以得出答案。

int numRookCaptures(char** board, int boardSize, int* boardColSize){
    int X, Y;
    int cnt = 0, i, j;
    for (i  = 0; i < 8; i++) {
        for (j = 0; j < 8; j++) {
            if (board[i][j] == 'R') {
                X = i;
                Y = j;
            }
        }
    }
    for (i = Y; i >= 0; i--) {
        if (board[X][i] == 'p') {
            cnt++;
            break;
        }
        if (board[X][i] == 'B') {
            break;
        }
    }
    for (i = Y; i < 8; i++) {
        if (board[X][i] == 'p') {
            cnt++;
            break;
        }
        if (board[X][i] == 'B') {
            break;
        }
    }
    for (i = X; i >= 0; i--) {
        if (board[i][Y] == 'p') {
            cnt++;
            break;
        }
        if (board[i][Y] == 'B') {
            break;
        }
    }
    for (i = X; i < 8; i++) {
        if (board[i][Y] == 'p') {
            cnt++;
            break;
        }
        if (board[i][Y] == 'B') {
            break;
        }
    }
    return cnt;
}