Note: The result may be very large, so you need to return a string instead of an integer. Credits: Special thanks to@tsfor adding this problem and creating all test cases. SOLUTION 1: 参考http://bookshadow.com/weblog/2015/01/13/leetcode-largest-number/的解法: 贪心思路:对于两个备选数...
原题地址: https://oj.leetcode.com/problems/largest-number/ 题目内容: Given a list of non negative integers, arrange them such that they form the largest number. For example, given[3, 30, 34, 5, 9], the largest formed number is9534330. Note: The result may be very large, so you n...
Largest Number 2. Solution 解析:这道题的关键在于想到需要编写一个比较函数来比较两个数字的“大小”,即两个数字排列的先后顺序。 Version 1 importfunctoolsclassSolution:deflargestNumber(self,nums):nums_str=[str(num)fornuminnums]nums_str=sorted(nums_str,key=functools.cmp_to_key(self.compare),reverse...
= bch: return ord(ach) - ord(bch) # 都相等,则两个字符串数字相同 return 0 # 将比较函数转换成 key cmp_key = cmp_to_key(cmp) class Solution: def kthLargestNumber(self, nums: List[str], k: int) -> str: #第 k 大就是第 n - k + 1 小 k = len(nums) - k + 1 # 按升序...
class Solution { public: string largestNumber(vector<int>& nums) { sort(nums.begin(), nums.end(), compare);//直接对数组进行自己定义的字典顺序“排序” string res; for (auto n : nums) res += to_string(n); auto pos = res.find_first_not_of('0');//找到第一个不为‘0’的位置 ...
Largest Number 二、解题 关键点就在于,如何对比两个数的大小?(理解为两个数谁应该放在前面),解法是按照顺序拼接两个字母串进行比较,如果a +b串 大于 b+a串,那么a比较大(即题意中理解的a应该放在前面),反之b比较大。 三、尝试与结果 classSolution:defsmaller(self,a,b):strA=str(a)+str(b)strB=str(...
class Solution { static bool cmp(string lhv, string rhv) { return (lhv+rhv) > (rhv+lhv); } public: string largestNumber(vector<int>& nums) { vector<string> numStrs; for(size_t i = 0; i < nums.size(); ++i) { numStrs.push_back(to_string(nums[i])); ...
## LeetCode 215E -fromtypingimportListclassSolution:deffindKthLargest(self,nums:List[int],k:int)->int:nums.sort(reverse=True)returnnums[k-1] 运行尝试: 提交到 LeetCode: 通过。 这个排序的写法还可以简化: ## LeetCode 215E -fromtypingimportListclassSolution:deffindKthLargest(self,nums:List[int...
Solution 1: Grouping + Equi-Join + Subquery We can first count the number of employees in each department, denoted as tableT. Then we joinTwith theEmployeestable, with the join condition beingT.dep_id = Employees.dep_idandEmployees.position = 'Manager'. This way, we can get the manager ...
Explanation: The digits that appear in s are [1]. There is no second largest digit. Constraints: 1 <= s.length <= 500 sconsists of only lowercase English letters and/or digits. 也可以用list或者直接用2个变量处理。 classSolution{publicintsecondHighest(String s){PriorityQueue<Integer>pq=newPr...