leecode每日一题:999. 车的可用捕获量

  • 2020 年 3 月 30 日
  • 筆記

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

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:

  • 题目的意思是中间R走一下能吃p的次数,方向是上下左右四个方向
  • 限制条件是走一次,以及遇到B就表明该方向不通
class Solution:      def numRookCaptures(self, board: List[List[str]]) -> int:          cnt, st, ed = 0, 0, 0          # 方向数组          direction = [(0, 1), (0, -1), (1, 0), (-1, 0)]          # 找到R          for i in range(8):              for j in range(8):                  if board[i][j] == 'R':                      st = i                      ed = j          # 朝着四个方向探索          for i in range(4):              step = 0              while True:                  dx, dy = direction[i]                  tx = st + step * dx                  ty = ed + step * dy                  if tx < 0 or tx >= 8 or ty < 0 or ty >= 8 or board[tx][ty] == 'B':                      break                  if board[tx][ty] == 'p':                      cnt += 1                      break                  step += 1  #每次移动的范围,第一次移动一格,找不到的话移动两格          return cnt