【PythonCheckiO题解】PawnBrotherhood

CheckiO 是面向初学者和高级程序员的编码游戏&＃xff0c;使用 Python 和 Javascript 解决棘手的挑战和有趣的任务&＃xff0c;从而提高你的编码技能&＃xff0c;本博客主要记录自己用 Python 在闯关时的做题思路和实现代码&＃xff0c;同时也学习学习其他大神写的代码。

CheckiO 官网&＃xff1a;https://checkio.org/

我的 CheckiO 主页&＃xff1a;https://py.checkio.org/user/TRHX/

CheckiO 题解系列专栏&＃xff1a;https://itrhx.blog.csdn.net/category_9536424.html

CheckiO 所有题解源代码&＃xff1a;https://github.com/TRHX/Python-CheckiO-Exercise

题目描述

【Pawn Brotherhood】&＃xff1a;这次的题目只有英文的&＃xff0c;而且有点长&＃xff0c;用百度翻译翻译了一下&＃xff0c;然而一团糟&＃xff0c;只知道是关于国际象棋的&＃xff0c;其他不知道说的啥&＃xff0c;为了搞懂题目意思&＃xff0c;我甚至还百度了一下国际象棋怎么下&＃xff0c;然而并没有什么用&＃xff0c;于是我用我那小学六年级水平的英语&＃xff0c;配合翻译软件和给的示例图&＃xff0c;搞了半天才弄清楚题目啥意思&＃xff0c;简单来说&＃xff0c;题目会给一个字典&＃xff0c;字典中的值就是一枚棋子的坐标&＃xff0c;我们要统计安全的棋子的个数&＃xff0c;怎么判断棋子是否安全呢&＃xff1f;如果棋子的左下角或者右下角有其他棋子&＃xff0c;那么这枚棋子就是安全的&＃xff0c;以左图的 b4 位置的棋子为例&＃xff0c;a3 位置没有棋子&＃xff0c;但是 c3 位置有棋子&＃xff0c;所以 b4 位置的棋子就是安全的&＃xff0c;对于 d2 位置的棋子&＃xff0c;因为 c1 和 e1 位置都没棋子&＃xff0c;所以就不安全。

【链接】&＃xff1a;https://py.checkio.org/mission/pawn-brotherhood/

【输入】&＃xff1a;由棋子坐标组成的字典&＃xff08;字典&＃xff09;

【输出】&＃xff1a;处于安全位置的棋子的个数&＃xff08;整数&＃xff09;

【前提】&＃xff1a;0

【范例】&＃xff1a;

safe_pawns &＃xff08;{
“ b4” &＃xff0c;“ d4” &＃xff0c;“ f4” &＃xff0c;“ c3” &＃xff0c;“ e3” &＃xff0c;“ g5” &＃xff0c;“ d2” } &＃xff09;&＃61;&＃61; 6
safe_pawns &＃xff08;{
“ b4” &＃xff0c;“ c4” &＃xff0c;“ d4” &＃xff0c;“ e4” &＃xff0c;“ f4” &＃xff0c;“ g4” &＃xff0c;“ e5” } &＃xff09;&＃61;&＃61; 1


解题思路

个人感觉这题还是比较简单的&＃xff0c;但是官网定的难度是中等&＃xff0c;直接将字典中每一个值转换成列表&＃xff0c;列表的元素分别是横向坐标和纵向坐标&＃xff0c;对于其左下角的坐标&＃xff0c;可以将其横向坐标和纵向坐标都减一&＃xff0c;右下角的坐标类似&＃xff0c;因为横向坐标是 a b c 的形式&＃xff0c;我们可以先用 ord() 方法将其转换成 ASCII 码&＃xff0c;做减法完成后再用 chr() 方法将其转换成 chr 类型&＃xff0c;然后判断这个坐标是否在字典里就行了

代码实现

def safe_pawns(pawns: set) -> int:
num &＃61; 0
for i in pawns:
left_pawns &＃61; chr(ord(list(i)[0])-1) &＃43; chr(ord(list(i)[1])-1)
right_pawns &＃61; chr(ord(list(i)[0])&＃43;1) &＃43; chr(ord(list(i)[1])-1)
if left_pawns in pawns or right_pawns in pawns:
num &＃43;&＃61; 1
return num
if __name__ &＃61;&＃61; &＃39;__main__&＃39;:
# These "asserts" using only for self-checking and not necessary for auto-testing
assert safe_pawns({
"b4", "d4", "f4", "c3", "e3", "g5", "d2"}) &＃61;&＃61; 6
assert safe_pawns({
"b4", "c4", "d4", "e4", "f4", "g4", "e5"}) &＃61;&＃61; 1
print("Coding complete? Click &＃39;Check&＃39; to review your tests and earn cool rewards!")


大神解答

大神解答 NO.1

def safe_pawns(pawns):
answer &＃61; 0
for pawn in pawns :
if chr(ord(pawn[0])-1)&＃43;str(int(pawn[1])-1) in pawns or chr(ord(pawn[0])&＃43;1)&＃43;str(int(pawn[1])-1) in pawns : answer &＃43;&＃61;1
return answer


大神解答 NO.2

def getdiags(pawn):
c, r &＃61; map(ord, pawn)
return chr(c - 1) &＃43; chr(r - 1), chr(c &＃43; 1) &＃43; chr(r - 1)
def safe_pawns(pawns):
return len([p for p in pawns if any(d in pawns for d in getdiags(p))])


大神解答 NO.3

def safe_pawns(pawns):
def is_safe(p):
file, rank &＃61; ord(p[0]), int(p[-1])
return (chr(file-1)&＃43;str(rank-1) in pawns or
chr(file&＃43;1)&＃43;str(rank-1) in pawns)
return sum(is_safe(p) for p in pawns)


大神解答 NO.4

def safe_pawns(pawns):
safe &＃61; lambda s: {
chr(ord(s[0]) - 1) &＃43; str(int(s[1]) - 1),
chr(ord(s[0]) &＃43; 1) &＃43; str(int(s[1]) - 1)}
return sum([bool(set(pawns).intersection(safe(p))) for p in pawns ])


大神解答 NO.5

# migrated from python 2.7
def get_defenders(pawn):
x, y &＃61; list(map(ord, pawn))
return &＃39;%c%c&＃39; % (x - 1, y - 1), &＃39;%c%c&＃39; % (x &＃43; 1, y - 1)
def safe_pawns(pawns):
return sum(not pawns.isdisjoint(get_defenders(x)) for x in pawns)