leetcode 999. 车的可用捕获量(Available Captures for Rook)

Posted 2021-02-15 zhanzq1

目录

• 题目描述：
• 示例 1：
• 示例 2：
• 示例 3：
• 解法：

题目描述：

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\'

---

解法：

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();
        int x = 0, y = 0;   // the rook position (x, y)
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(board[i][j] == \'R\'){
                    x = i;
                    y = j;
                }
            }
        }
        // cout<<x<<", "<<y<<endl;
        int res = 0;
        int _x = 0, _y = 0;
        // up
        _x = x - 1;
        _y = y;
        while(_x >= 0 && board[_x][_y] == \'.\'){
            _x--;
        }
        if(_x >= 0 && board[_x][_y] == \'p\'){
            res++;
        }
        
        // down
        _x = x + 1;
        _y = y;
        while(_x < m && board[_x][_y] == \'.\'){
            _x++;
        }
        if(_x < m && board[_x][_y] == \'p\'){
            res++;
        }
        
        // left
        _x = x;
        _y = y - 1;
        while(_y >= 0 && board[_x][_y] == \'.\'){
            _y--;
        }
        if(_y >= 0 && board[_x][_y] == \'p\'){
            res++;
        }
        
        // right
        _x = x;
        _y = y + 1;
        while(_y < n && board[_x][_y] == \'.\'){
            _y++;
        }
        if(_y < n && board[_x][_y] == \'p\'){
            res++;
        }
        return res;
    }
};