999、车的可用捕获量

leetcode

Created At : 2020-07-26 00:19

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999、车的可用捕获量
题解
1. 1、找到车的位置，然后按照上、下、左、右四个方向找卒。

999、车的可用捕获量

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：

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

链接：https://leetcode-cn.com/problems/available-captures-for-rook

题解

最多可以捕获4个卒pawn。

1、找到车的位置，然后按照上、下、左、右四个方向找卒。

class Solution {
    public int numRookCaptures(char[][] board) {
        int N = board.length;
        int x = 0,y = 0;
        for(int i = 0;i < N;i++) {
            for(int j = 0;j < N;j++) {
                if (board[i][j] == 'R') {
                     x = i;
                     y = j;
                     break;
                }
            }
            if (board[x][y] == 'R') {
                break;
            }
        }
        int count = 0;
        if (canGetPawn(board,x,y,1,0)){
            count++;
        }
        if (canGetPawn(board,x,y,-1,0)) {
            count++;
        }
        if (canGetPawn(board,x,y,0,1)) {
            count++;
        }
        if (canGetPawn(board,x,y,0,-1)) {
            count++;
        }

        return count;
    }

    private boolean canGetPawn(char[][] board,int x,int y,int accx,int accy){
        boolean res = false;
        int i = x + accx;
        int j = y + accy;
        int N = board.length;
        while(i >= 0 && i < N && j >= 0 && j < N){
            if (board[i][j] == 'B') {
                break;
            }
            if (board[i][j] == 'p') {
                res = true;
                break;
            }
            i += accx;
            j += accy;
        }
        return res;
    }
}
// 另一种实现
class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int cnt = 0, st = 0, ed = 0;
        int dx[4] = {0, 1, 0, -1};
        int dy[4] = {1, 0, -1, 0};
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (board[i][j] == 'R') {
                    st = i;
                    ed = j;
                    break;
                }
            }
            if (board[et][ed] == 'R') {
                break;
            }
        }
        for (int i = 0; i < 4; ++i) {
            for (int step = 0;; ++step) {
                int tx = st + step * dx[i];
                int ty = ed + step * dy[i];
                if (tx < 0 || tx >= 8 || ty < 0 || ty >= 8 || board[tx][ty] == 'B') break;
                if (board[tx][ty] == 'p') {
                    cnt++;
                    break;
                }
            }
        }
        return cnt;
    }
};

// 作者：LeetCode-Solution
// 链接：https://leetcode-cn.com/problems/available-captures-for-rook/solution/che-de-ke-yong-bu-huo-liang-by-leetcode-solution/

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