比较版本号_

Python3

lintoon

class Solution:

def compare(self , version1: str, version2: str) -> int:

# write code here

#两个字符串列表化，同时将元素化为整形

x=version1.split('.')

y=version2.split('.')

x=[int(a) for a in x]

y=[int(b) for b in y]

#初始化循环的结束条件

z=min(len(x),len(y))

#初始化两个列表的初始索引为0

ct=0

#通过循环对两个列表的元素进行比较

while ct<z:

if x[ct]>y[ct]:

return 1

elif x[ct]<y[ct]:

return -1

#两个列表的同索引元素相等时，索引加一，继续循环

else:ct+=1

#循环结束仍未输出结果继续对列表长度进行判断

else:

#长度相等则两个版本号等价

if len(x)==len(y):

return 0

#长度不等且更长的列表超出的元素相加不为零，则更长的版本号更大

elif len(x)>len(y) and sum(x[len(y):])!=0:

return 1

elif len(x)<len(y) and sum(y[len(x):])!=0:

return -1

#存在长度不等，但超出部分的元素相加为零，实则两个版本号等价

else:return 0

发表于 2024-04-09 11:27:28 回复(0)

Python3

猛牛向前冲

def compare(self , version1: str, version2: str) -> int:
        version1 = version1.rstrip('.0')
        version2 = version2.rstrip('.0')
        list1 = version1.split('.')
        list2 = version2.split('.')
        ret = 0
        for index in range(len(list1)):
            num1 = list1[index]
            if index >= len(list2):
                ret = 1
                break
            num2 = list2[index]
            if num1== num2:
                continue
            if int(num1) > int(num2):
                ret = 1
                break
            elif int(num1) == int(num2):
                continue
            ret = -1
            break
        if ret == 0:
            if len(list1) < len(list2):
                ret = -1
        return ret

发表于 2023-10-16 15:42:57 回复(0)

Python3

可道非常道

class Solution:
    def compare(self , version1: str, version2: str) -> int:
        # write code here
        ver1 = version1.split('.')
        ver2 = version2.split('.')
        n1, n2 = len(ver1), len(ver2)
        
        for i in range(max(n1, n2)):
            num1 = int(ver1[i]) if i < n1 else 0
            num2 = int(ver2[i]) if i < n2 else 0
            
            if num1 > num2:
                return 1
            elif num1 < num2:
                return -1
        
        return 0

发表于 2023-10-16 09:53:26 回复(0)

Python3

RIGHTRICE

class Solution:
    def compare(self , version1: str, version2: str) -> int:
        version1_SP = version1.split('.')
        version2_SP = version2.split('.')
        while len(version1_SP)<len(version2_SP):
            version1_SP.append(0)
        while len(version1_SP)>len(version2_SP):
            version2_SP.append(0)

        for ii in range(len(version2_SP)):
            if int(version1_SP[ii])<int(version2_SP[ii]):
                return -1
            elif int(version1_SP[ii])>int(version2_SP[ii]):
                return 1
        return 0

发表于 2023-03-01 09:58:32 回复(0)

Python3

Vivimore

时间复杂度：O(n) 空间复杂度O(1)

class Solution:
    def compare(self , version1: str, version2: str) -> int:
        v1 = list(map(int, version1.split('.')))       
        v2 = list(map(int, version2.split('.')))
        if len(v1) > len(v2):
            v2.extend([0 for i in range(len(v1) - len(v2))])
        elif len(v1) < len(v2):
            v1.extend([0 for i in range(len(v2) - len(v1))])
        for i in range(len(v1)):
            if v1[i] > v2[i]:
                return 1
            elif v1[i] < v2[i]:
                return -1
        return 0

class Solution:
    def compare(self , version1: str, version2: str) -> int:
        n1 = len(version1)
        n2 = len(version2)
        i, j = 0, 0
        while i < n1 or j < n2:
            num1 = 0
            while i < n1 and version1[i] != '.':
                num1 = num1 * 10 + int(version1[i])
                i += 1
            i += 1
            num2 = 0
            while j < n2 and version2[j] != '.':
                num2 = num2 * 10 + int(version2[j])
                j += 1
            j += 1
            if num1 < num2:
                return -1
            elif num1 > num2:
                return 1
        return 0

发表于 2022-05-16 16:24:49 回复(0)

Python3

Andronie

class Solution:
    def compare(self , version1: str, version2: str) -> int:
        # write code here
        v1 = version1.split('.')
        v2 = version2.split('.')
        if len(v1) <= len(v2):
            res = self.c(v1, v2)
        elif len(v1) > len(v2):
            res = - self.c(v2, v1)
        return res
    def c(self, a, b):
        for e, i in enumerate(a):
            if int(i) > int(b[e]):
                return 1
            elif int(i) < int(b[e]):
                return -1
        for j in b[e + 1:]:
            if int(j) != 0:
                return -1
        return 0

发表于 2021-12-21 17:00:55 回复(0)

比较版本号

输入

输出

说明

输入

输出

说明

输入

输出

说明

输入

输出

说明

输入

输出

说明

问题信息

热门推荐

通过挑战的用户

相关试题