165-171 我直接送一道 好家伙

166: 比较版本号 

用split分割 用int比较

class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        v1 = version1.split('.')
        v2 = version2.split('.')
        n1 = len(v1)
        n2 = len(v2)
        n = min(n1,n2)
        i = 0
        while i < n :
            if int(v1[i]) > int(v2[i]):
                return 1
            elif int(v1[i]) < int(v2[i]):
                return -1
            i += 1
        if i == n1:
            for i in range(n,n2):
                if int(v2[i]) != 0:
                    return -1
        if i == n2:
            for i in range(n,n1):
                if int(v1[i]) != 0:
                    return 1
        return 0

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/compare-version-numbers/solution/yong-splitfen-ge-yong-intbi-jiao-by-yizh-jzre/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

166 :分数到小数 :

我直接CV工程师  因为不知道半途开始的循环咋办  原来是模仿除法 

class Solution:
    def fractionToDecimal(self, numerator: int, denominator: int) -> str:
        if numerator % denominator == 0:
            return str(numerator // denominator)

        s = []
        if (numerator < 0) != (denominator < 0):
            s.append('-')

        # 整数部分
        numerator = abs(numerator)
        denominator = abs(denominator)
        integerPart = numerator // denominator
        s.append(str(integerPart))
        s.append('.')

        # 小数部分
        indexMap = {}
        remainder = numerator % denominator
        while remainder and remainder not in indexMap:
            indexMap[remainder] = len(s)
            remainder *= 10
            s.append(str(remainder // denominator))
            remainder %= denominator
        if remainder:  # 有循环节
            insertIndex = indexMap[remainder]
            s.insert(insertIndex, '(')
            s.append(')')

        return ''.join(s)

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/fraction-to-recurring-decimal/solution/cvgong-cheng-shi-by-yizhu-jia-umwa/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

　　167: 输入有序数组 

好像二分更好   但是双指针 你值得拥有  

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        left = 0
        right = len(numbers) -1
        while left < right:
            if numbers[left] + numbers[right] == target:
                return [left+1,right+1]
            elif numbers[left] + numbers[right] > target:
                right -= 1
            else:
                left += 1


作者：yizhu-jia
链接：https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted/solution/you-er-fen-fa-ma-wo-zen-mo-bu-zhi-dao-by-rsct/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

168 :excel 列表名称 

简单题 

class Solution:
    def convertToTitle(self, columnNumber: int) -> str:
        rel = ''
        while columnNumber != 0 :
            columnNumber -= 1
            rel = ''.join((chr(columnNumber%26+65),rel))
            columnNumber = columnNumber//26
        return rel

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/excel-sheet-column-title/solution/jian-1jian-de-jing-sui-a-by-yizhu-jia-krly/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

169 : 

多数元素 因为之前见过 所以直接写了  计数到0就换数 

class Solution:
    def majorityElement(self, nums: List[int]) -> int:

        curnum =  nums[0]
        count = 0
        for each in nums:
            if curnum == each:

                count += 1
            else:
                if count == 0:
                    curnum = each
                else:
                    count -= 1
        return curnum

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/majority-element/solution/ji-shu-dao-0huan-shu-by-yizhu-jia-7ufh/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

171: excel表列序号 

简单题 

class Solution:
    def titleToNumber(self, columnTitle: str) -> int:
        rel = 0
        for each in columnTitle:
            rel = rel * 26 + ord(each) - ord('A') + 1
        return rel


作者：yizhu-jia
链接：https://leetcode-cn.com/problems/excel-sheet-column-number/solution/zhe-ti-da-jia-fang-fa-ying-gai-du-yi-yan-lmak/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。