【LeetCode】999. 车的可用捕获量

题目

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'

思路

每次可以在四个方向上分别移动一次,先找出'R'位置,然后在四个方向上移动。

代码

时间复杂度:O(n^2)
空间复杂度:O(1)

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        vector<int> dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0};
        int tx, ty, res = 0;
        //1. 找出R位置
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (board[i][j] == 'R') {
                    tx = i;
                    ty = j;
                    break;
                }
            }
        }
        //2. 分别朝四个方向上移动
        for (int i = 0; i < 4; ++i) {
            int step = 0;
            while (step < 8) {
                int x = tx + step * dx[i];
                int y = ty + step * dy[i];
                if (x < 0 || x >= 8 || y < 0 || y >= 8 || board[x][y] == 'B') break;
                if (board[x][y] == 'p') {
                    ++res;
                    break;
                }
                ++step;
            }
        }

        return res;

    }
};
posted @ 2020-03-26 22:22  Galaxy_hao  阅读(86)  评论(0编辑  收藏  举报