支持sum(2,3)和sum(2)(3)语法

js中sum(2,3,4)和sum(2)(3)(4)都返回9并要求扩展性

网上有很多关于sum(1)(2)(3),sum(1,2,3)之类的面试题要求输出相同的结果6并要求可以满足扩展,即有多个参数时也能符合题设的要求,所以自己写了部分例子可以大概满足这些面试题的要求 <!DOCTYPE html><html lang="en"><head> <meta charset="UTF-8"> <title>Document</title></head><

Subarray Sum & Maximum Size Subarray Sum Equals K

Subarray Sum Given an integer array, find a subarray where the sum of numbers is zero. Your code should return the index of the first number and the index of the last number. Notice There is at least one subarray that it's sum equals to zero. Example

编写一个求和函数sum，使输入sum(2)(3)或输入sum(2,3)，输出结果都为5

昨天的笔试题,做的一塌糊涂,题目考的都很基础而且很细,手写代码对我来说是硬伤啊.其中有一道是这个,然而看到题目的时候,根本没有想到arguments:然后现在就恶补一下. arguments:用在函数内部,表示传给该函数的所有的实参的集合. 虽然arguments对象并不是一个数组,但是访问单个参数的方式与访问数组元素的方式相同.例如:arguments[0],arguments[1]...arguments[n]. function foo(){ var sum = 0; console.lo

leetcode 112. Path Sum 、 113. Path Sum II 、437. Path Sum III

112. Path Sum 自己的一个错误写法: class Solution { public: bool hasPathSum(TreeNode* root, int sum) { if(root == NULL) return false; ; return hasPathSum(root,sum,value); } bool hasPathSum(TreeNode* root,int sum,int value){ if(root == NULL){ if(value == sum) r

Subarray Sum & Maximum Size Subarray Sum Equals K && Subarray Sum Equals K

Subarray Sum Given an integer array, find a subarray where the sum of numbers is zero. Your code should return the index of the first number and the index of the last number. Notice There is at least one subarray that it's sum equals to zero. Example

[LeetCode] Range Sum Query - Immutable & Range Sum Query 2D - Immutable

Range Sum Query - Immutable Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. Example: Given nums = [-2, 0, 3, -5, 2, -1] sumRange(0, 2) -> 1 sumRange(2, 5) -> -1 sumRange(0, 5) -> -3 Note: You

ORACEL12c求和SUM(a+b)与SUM(a)+SUM(b)结果不一致问题

今天在oracle12c数据库遇到的问题,下面重现一下: 1.先创建一个临时表 temp.一个数据表bitest_sum; -- Create table create table PEC.BITEST_SUM ( yyyymm ), factno ), dept_no ), product ), amt not null, amt_ys not null ) tablespace BITBS; -- Add comments to the table comment on table PEC.

leetcode 39. Combination Sum 、40. Combination Sum II 、216. Combination Sum III

39. Combination Sum 依旧与subsets问题相似,每次选择这个数是否参加到求和中因为是可以重复的,所以每次递归还是在i上,如果不能重复,就可以变成i+1 class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> result; vector

39. Combination Sum + 40. Combination Sum II + 216. Combination Sum III + 377. Combination Sum IV

▶ 给定一个数组和一个目标值.从该数组中选出若干项(项数不定),使他们的和等于目标值. ▶ 36. 数组元素无重复 ● 代码,初版,19 ms .从底向上的动态规划,但是转移方程比较智障(将待求数分解为左右两个半段,分别找解,拼在一起,再在接缝上检查是否是重复解). class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

leetcode 1.Two Sum 、167. Two Sum II - Input array is sorted 、15. 3Sum 、16. 3Sum Closest 、 18. 4Sum 、653. Two Sum IV - Input is a BST

1.two sum 用hash来存储数值和对应的位置索引,通过target-当前值来获得需要的值,然后再hash中寻找错误代码1: Input:[3,2,4]6Output:[0,0]Expected:[1,2] 同一个数字不能重复使用,但这个代码没排除这个问题 class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> result; uno

29.Combination Sum（和为sum的组合）

Level: Medium 题目描述: Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may be chosen from

sum(sum(abs(y))) 中 sum(sum())什么意思?

>> y=[1 3;2 5] y = 1 3 2 5 >> sum(y) ans = 3 8 >> sum(sum(y)) ans = 11 -------------------------------------------------------------- 即: sum(sum(y)) 求矩阵y所有元素的和. sum(y) 是对矩阵各个列求和

写一个 sum方法，在使用下面任一语法调用时，都可以正常工作

console.log(sum(2,3)); // Outputs 5 console.log(sum(2)(3)); // Outputs 5 (至少)有两种方法可以做到: 方法1: function sum(x) { if (arguments.length == 2) { return arguments[0] + arguments[1]; } else { return function(y) { return x + y; }; } } 在JavaScript中,函数可以提供到 ar

leetcode@ [303/304] Range Sum Query - Immutable / Range Sum Query 2D - Immutable

https://leetcode.com/problems/range-sum-query-immutable/ class NumArray { public: vector<int> vec; public: NumArray(vector<int> &nums) { if(nums.empty()) return; else { vec.push_back(nums[]); ;i<nums.size();++i) vec.push_back(vec[i-] +

Project Euler Problem 18-Maximum path sum I & 67-Maximum path sum II

基础的动态规划...数塔取数问题. 状态转移方程: dp[i][j] = num[i][j] + max(dp[i+1][j],dp[i+1][j+1]);

记一个同时支持模糊匹配和静态推导的Atom语法补全插件的开发过程: 序

简介过去的一周,都睡的很晚,终于做出了Atom上的APICloud语法提示与补全插件:apicloud_autocomplete.个中滋味,感觉还是有必要记录下来的.代码基于 GPL-3.0 开源,所以我可以较为详细的记录一些很难被理解和体会的技术细节.APICloud目前已有Studio,VSCode,Webstrom和Sublime的语法补全插件,但是毫无疑问,我做的这款,是目前为止最好的 -- 唯一的一个支持100%所有API,唯一的一个同时支持模糊匹配和静态推导语法提示插件! 可能你会

linq 对Sum()函数的支持

首先看一段SQL语句: SELECT SUM(TASKAPPR) AS APPRCOUNT, SUM(TASKLOCKED) AS LOCKEDCOUNT, SUM(TASKCHECKED) AS CHECKEDCOUNT FROM ( SELECT ( END) AS TASKAPPR, ( END) AS TASKLOCKED, ( END) AS TASKCHECKED FROM TASK) AS TEMP 上面是对一个叫Task表的三个bit类型的字段进行数据统计,用到了三个case w

[LeetCode] Range Sum Query - Mutable 区域和检索 - 可变

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index i to val. Example: Given nums = [1, 3, 5] sumRange(0, 2) -> 9 update(1, 2

【BZOJ-3638&3272&3267&3502】k-Maximum Subsequence Sum 费用流构图 + 线段树手动增广

3638: Cf172 k-Maximum Subsequence Sum Time Limit: 50 Sec Memory Limit: 256 MBSubmit: 174 Solved: 92[Submit][Status][Discuss] Description 给一列数,要求支持操作: 1.修改某个数的值 2.读入l,r,k,询问在[l,r]内选不相交的不超过k个子段,最大的和是多少. Input The first line contains integer n (1 ≤ n

[Swift]LeetCode113. 路径总和 II | Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 Return: [ [5,4,

[Swift]LeetCode209. 长度最小的子数组 | Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead. Example: Input: s = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: the subarr

支持sum(2,3)和sum(2)(3)语法

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