leetcode---999. 車的可用捕獲量

今天的題也很簡單，感覺這幾天leetcode的每日一題有點水…
題目鏈接----->999. 車的可用捕獲量

方法一：掃描其所在的行和列

int numRookCaptures(vector<vector<char>>& board) {
    int len = board.size();
    int i,j,x,y,ans = 0,sum = 0;
    for(i = 0; i < len;i++)
    {
        for(j = 0;j < len;j++)
        {
            if(board[i][j] == 'R')
            {
                x = i;
                y = j;
                break;
            }
        }
        if(j < len)
            break;
    }
    for(i = 0; i < len ;i++)
    {
        if(i > y)
        {
            if(board[x][i] == 'B')
                break;
            else if(board[x][i] == 'p')
            {
                ans += 1;
                break;
            }
        }
        if(board[x][i] == 'p')
            sum = 1;
        else if(board[x][i] == 'B')
            sum = 0;
        else if(board[x][i] == 'R')
            ans += sum;
    }
    sum = 0;
    for(j = 0;j < len;j++)
    {
        if(j > x)
        {
            if(board[j][y] == 'B')
                break;
            else if(board[j][y] == 'p')
            {
                ans += 1;
                break;
            }
        }
        if(board[j][y] == 'p')
            sum = 1;
        else if(board[j][y] == 'B')
            sum = 0;
        else if(board[j][y] == 'R')
            ans += sum;
    }
    return ans;
}

方法二：模擬

int numRookCaptures(vector<vector<char>>& board) {
    int len = board.size();
    int x,y,ans = 0;
    int fxy[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
    for(int i = 0;i < len;i++)
    {
        for(int j = 0; j < len;j++)
        {
            if(board[i][j] == 'R')
            {
                x = i;
                y = j;
                break;
            }
        }
    }
    for(int i = 0;i < 4;i++)
    {
        int dx = x;
        int dy = y;
        while(1)
        {
            dx += fxy[i][0];
            dy += fxy[i][1];
            if(dx < 0 || dx >= len || dy < 0 || dy >= len || board[dx][dy] == 'B')
                break;
            if(board[dx][dy] == 'p')
            {
                ans++;
                break;
            }
        }
    }
    return ans;
}