车的可用捕获量（3.26leetcode每日打卡）

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。

 

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3

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。

 

思路：题目很简单就是简单的遍历四个方向即可，主要是好好审题，比如抓到一个卒就可停止，好好审题方能事半功倍。

int numRookCaptures(char** board, int boardSize, int* boardColSize)
{
    int cnt = 0;
    int row,col;
    int i,j;
    
    //搜寻白色车所在位置
    //车 R
    //象 B
    //卒 p
    //空方格 .
    for(i = 0; i < 8; i++)
    {
        for(j = 0; j < 8; j++)
        {
            if(board[i][j] == 'R')
            {
                row = i;
                col = j;
                break;
            }
        }
    }
    
    //搜寻右边
    for(i = col+1; i < 8; i++)
    {
        if(board[row][i] == 'p')
        {
            cnt++;
            break;
        }
        if(board[row][i] == 'B')
        break;
    }

    //搜寻左边
    for(i = col-1; i >= 0; i--)
    {
        if(board[row][i] == 'p')
        {
            cnt++;
            break;
        }
        if(board[row][i] == 'B')
        break;
    }

    //搜寻下边
    for(i = row+1; i < 8; i++)
    {
        if(board[i][col] == 'p')
        {
            cnt++;
            break;
        }
        if(board[i][col] == 'B')
        break;      
    }

    //搜寻上边
    for(i = row-1; i >= 0; i--)
    {
        if(board[i][col] == 'p')
        {
            cnt++;
            break;
        }
        if(board[i][col] == 'B')
        break;      
    }
    return cnt;
}