代码(Python3) def cmp(a: str, b: str) -> int: """ 比较两个数字字符串的大小 """ # 如果长度不等,则长度更长的数更大 if len(a) != len(b): return len(a) - len(b) # 如果长度相等,则比较字符串的大小 for ach, bch in zip(a, b): # 对应位置字符不等,则数字大的数更大 if...
Avoid usingif-stmtsto find maximum. Use python builtinmax. Use either generator orfilterto find only the odd numbers. Using builtins like this is safer/more reliable because it is simpler to compose them, the code is well-tested, and the code executes mostly in C (rather than multiple by...
print('This program will find the largest odd number among the three entered.\nSo, let\'s start...\n') x = int(input('Enter the 1st number: ')) y = int(input('Enter the 2nd number: ')) z = int(input('Enter the 3rd number: ')) strToPrint = ' is the largest odd number....
if '0' not in digit_str[i: i+n]: prod_ = reduce(lambda x,y: int(x) * int(y), digit_str[i: i+n]) else: i = digit_str.index('0', i) + 1 continue else: if digit_str[i+n-1] == '0': prod_12 = 0 i += n continue else: prod_ = prod_12 * int(digit_str[i...
master LintCodeInPython/largest-triangle-area.py / Jump to Go to file 22 lines (21 sloc) 896 Bytes Raw Blame class Solution: """ @param points: List[List[int]] @return: return a double """ def largestTriangleArea(self, points):...
LargestRectangle in Histogram http://www.cnblogs.com/zuoyuan/p/3783993.html https://shenjie1993.gitbooks.io/leetcode-python/084%20Largest 58120 LargestNumber ), reverse=True) return str(int(''.join(nums_str))) Reference https://leetcode.com/problems/largest-number ...
The project is mainly crawling the largest adult site int the world -- PornHub,contains video title、 duration、 mp4 link、 cover url and specific PornHub link. Project crawling PornHub.com with simple structure and fastly. Crawling PornHub video speed can reach 5 million / day or more. Specif...
每次返回数组中的最后一个元素即可,该元素就是第k个最大的元素 代码如下: classKthLargest:def__init__(self,k:int,nums:List[int]):self.k=kiflen(nums)>=self.k:self.init_list=sorted(nums,reverse=True)[:self.k]else:self.init_list=sorted(nums,reverse=True)defadd(self,val:int)->int:index...
Every nums[i] will be an integer in the range [0, 99]. 思路1: 非排序,如果存在一个数,比其他任何数的两倍还大必然是最大数。时间复杂度为O(n2) Java版本: public int dominantIndex(int[] nums) { int n = nums.length; for (int i = 0; i < n; ++i) { ...
class Solution(object): def findKthLargest(self, nums, k): """ :type nums: List[int] :type k: int :rtype: int """ for i in range(k - 1): nums.remove(max(nums)) return max(nums) 方法二:排序排序之后直接找倒数第k个数字即可。class Solution { public: int findKthLargest(vector<...