通過乙個函式來查詢陣列中所有的字串最長公共字首。(第乙個是我寫的,後面有幾個很經典的,看不懂請看力扣官網)
class solution(object):
def longestcommonprefix1(self, strs):
""":type strs: list[str]
:rtype: str
"""length = len(strs)
if not length:
return ""
i = 0
prepare_str = strs[0]
inner_length = len(min(strs, key=len))
while i < inner_length:
flag = true # 判斷是否發現了不一樣的字串
for j in range(1, length):
if strs[j][i] != prepare_str[i]:
flag = false
break
if not flag or i + 1 > inner_length:
break
i += 1
return prepare_str[:i]
def longestcommonprefix2(self, strs):
"""這個方法很好,先將和第二個字串相同的留下,每次會是的substr長度減少
:type strs: list[str]
:rtype: str
"""if len(strs) == 0:
return ""
strs = sorted(strs)
substr = strs[0]
for i in range(1, len(strs)):
while strs[i].find(substr) != 0:
substr = substr[:-1]
if substr == "":
return substr
return substr
def longestcommonprefix3(self, strs):
"""說實話看不懂
:type strs: list[str]
:rtype: str
"""def lcp(start, end):
if start == end:
return strs[start]
mid = (start + end) // 2
lcpleft, lcpright = lcp(start, mid), lcp(mid + 1, end)
minlength = min(len(lcpleft), len(lcpright))
for i in range(minlength):
if lcpleft[i] != lcpright[i]:
return lcpleft[:i]
return lcpleft[:minlength]
return "" if not strs else lcp(0, len(strs) - 1)
def longestcommonprefix4(self, strs):
s=""
zip_str = zip(*strs)
for i in zip_str:
if len(set(i)) == 1:
s += i[0]
else:
return s
return s
def longestcommonprefix5(self, strs) -> str:
if not strs:
return ""
length, count = len(strs[0]), len(strs)
for i in range(length):
c = strs[0][i]
if any(i == len(strs[j]) or strs[j][i] != c for j in range(1, count)):
return strs[0][:i]
return strs[0]
def longestcommonprefix6(self, strs) -> str:
def lcp(start, end):
if start == end:
return strs[start]
mid = (start + end) // 2
inner_left, inner_right = lcp(start, mid), lcp(mid + 1, end)
min_len = min(len(inner_left), len(inner_right))
for i in range(min_len):
if inner_left[i] != inner_right[i]:
return inner_left[:i]
return inner_left[:min_len]
left, right = 0, len(strs) - 1
return "" if not strs else lcp(left, right)
def longestcommonprefix(self, strs) -> str:
"""二分法查詢還可以這樣用,很經典"""
def is_common_prefix(length):
str0, count = strs[0][:length], len(strs)
return all(strs[i][:length] == str0 for i in range(1, count))
if not strs:
return ""
min_length = min(len(s) for s in strs)
low, high = 0, min_length
while low < high:
mid = (high - low + 1) // 2 + low
if is_common_prefix(mid):
low = mid
else:
high = mid - 1
return strs[0][:low]
if __name__ == '__main__':
s = solution()
strs = ["flower", "flow", "flight"]
# strs = ["dog", "d","dacecar", "d"]
# strs = [""]
print(s.longestcommonprefix(strs))
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