every blog every motto: You will never know unless you try
0. 前言
生活好难,再坚持坚持!
1. 字符串
1.1 题目
1.2
1.2.1 方法一:
class Solution:
def getHint(self, secret: str, guess: str) -> str:
# 方法一:
bulls = 0 # 公牛
cows = 0 # 奶牛
bulls_cows = 0 # 公牛+母牛
# 遍历找位置相同,值相同的个数
for i in range(len(secret)):
if secret[i] == guess[i]:
bulls += 1
unqiue = set(guess) # 去重
for ele in unqiue:
if ele in secret:
bulls_cows += min(guess.count(ele), secret.count(ele))
cows = bulls_cows-bulls
return str(bulls)+'A'+str(cows)+'B'
1.2.2 方法二:
class Solution:
def getHint(self, secret: str, guess: str) -> str:
secret_count = [0] * 10 # 统计个数
guess_count = [0] * 10
bulls = 0 # 公牛
cows = 0 # 奶牛
for index in range(len(guess)):
# 公牛,+1
if secret[index] == guess[index]:
bulls += 1
# 非公牛,统计数字
else:
# 当前这个数字在secret中个数+1
secret_count[ord(secret[index]) - ord('0')] += 1
# 同理,guess统计数字中+1
guess_count[ord(guess[index]) - ord('0')] += 1
# 相同数字取最小,即为奶牛,因为已经考虑公牛了,所以这里不需要考虑公牛
for index in range(10):
cows += min(secret_count[index],guess_count[index])
return str(bulls)+'A'+str(cows)+'B'
1.2.3
class Solution:
def getHint(self, secret: str, guess: str) -> str:
# --------------------------------------------------------
# 方法三:
# 统计非公牛在secret和guess中的字符个数
unmatch_count = [0] * 10
bulls = 0 # 公牛
cows = 0 # 奶牛
for index in range(len(guess)):
# 判断是否为公牛,是+1
if secret[index] == guess[index]:
bulls += 1
# 非公牛,该数放在统计数组中计数,secret 为正,guess为负。
else:
# 遍历到secret中的这个数在统计数组中为负,则表明之前在guess中出现过,奶牛+1
if unmatch_count[ord(secret[index]) - ord('0')] < 0:
cows += 1
# 遍历到secret中的这个数,在统计数组中+1
unmatch_count[ord(secret[index]) - ord('0')] += 1
# 遍历到guess中的这个在统计数组中为正,则表明之前在secret中出现过,奶牛+1
if unmatch_count[ord(guess[index]) - ord('0')] > 0:
cows += 1
unmatch_count[ord(guess[index]) - ord('0')] -= 1
return str(bulls) + 'A' + str(cows) + 'B'