72 lines
1.3 KiB
Markdown
72 lines
1.3 KiB
Markdown
---
|
|
title: leetcode-999
|
|
date: 2020-03-30 21:03:25
|
|
tags:
|
|
categories: leetcode
|
|
---
|
|
|
|
|
|
### 999. 可以被一步捕获的棋子数
|
|
|
|
[题目](https://leetcode-cn.com/problems/available-captures-for-rook/)
|
|
|
|
|
|
|
|
<!--more-->
|
|
|
|
|
|
|
|
遍历一遍找到车的坐标，然后按上下左右四个方向循环一下看碰到的第一个棋子是什么。
|
|
|
|
|
|
|
|
```python
|
|
class Solution:
|
|
def numRookCaptures(self, board) -> int:
|
|
i = j = 0
|
|
for row in board:
|
|
if 'R' in row:
|
|
break
|
|
i += 1
|
|
j = row.index('R')
|
|
count = 0
|
|
# right
|
|
for x in range(j + 1, 8):
|
|
if row[x] == 'p':
|
|
count += 1
|
|
break
|
|
if row[x] == 'B':
|
|
break
|
|
# left
|
|
for x in range(j, 0, -1):
|
|
if row[x] == 'p':
|
|
count += 1
|
|
break
|
|
if row[x] == 'B':
|
|
break
|
|
# up
|
|
for x in range(i, 0, -1):
|
|
if board[x][j] == 'p':
|
|
count += 1
|
|
break
|
|
if board[x][j] == 'B':
|
|
break
|
|
# down
|
|
for x in range(i+1, 8):
|
|
if board[x][j] == 'p':
|
|
count += 1
|
|
break
|
|
if board[x][j] == 'B':
|
|
break
|
|
|
|
return count
|
|
|
|
#36 ms 13.6 MB
|
|
|
|
```
|
|
|
|
|
|
|
|
问题不难，官方解答中给了一个方向数组的概念，上下左右是 (0, 1) (0, -1) (-1, 0) (1, 0)，有点像向量的意思。走的路线等于方向数组乘以步数。
|
|
|