leetcode--999. 车的可用捕获量

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’

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        //1.找到R
        //2.北，西，南，东，停止条件：吃到p，遇到B，或到边缘
        int i = 0, j = 0, ans = 0;
        for (; i<8; ++i) {
            for (j=0; j<8; ++j) {
                if (board[i][j] == 'R') {
                    for (int k=i; k>=0; k--) {
                        if ('B' == board[k][j]) break;
                        if ('p' == board[k][j]) {
                            ans++;
                            break;
                        }
                    }
                    for (int k=i; k<8; k++) {
                        if ('B' == board[k][j]) break;
                        if ('p' == board[k][j]) {
                            ans++;
                            break;
                        }
                    }
                    for (int k=j; k>=0; k--) {
                        if ('B' == board[i][k]) break;
                        if ('p' == board[i][k]) {
                            ans++;
                            break;
                        }
                    }
                    for (int k=j; k<8; k++) {
                        if ('B' == board[i][k]) break;
                        if ('p' == board[i][k]) {
                            ans++;
                            break;
                        }
                    }
                }
            }
        }
        return ans;
    }
};
/*0ms,6.3MB*/

时间复杂度：O(N^2)
空间复杂度：O(1)

优化版：

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        //1.找到R
        //2.北，西，南，东，停止条件：吃到p，遇到B，或到边缘
        int dx[] = {-1, 1, 0, 0};
        int dy[] = {0, 0, -1, 1};
        for (int i=0; i<8; ++i) {
            for (int j=0; j<8; ++j) {
                if (board[i][j] == 'R') {
                    int ans = 0;
                    for (int k=0; k<4; k++) {
                        int x = i, y = j;
                        while (true) {
                            x += dx[k];
                            y += dy[k];
                            if(x<0 || x>=8 || y<0 || y>=8 || board[x][y] == 'B') break;
                            if (board[x][y] == 'p') {
                                ans++;
                                break;
                            }
                        }
                    }
                    return ans;
                }
            }
        }
        return 0;
    }
};
/*0ms,6.1MB*/

参考链接：https://leetcode-cn.com/problems/available-captures-for-rook/solution/jian-dan-java100-by-sweetiee/