sum()的参数是一个list: >>> sum([1,2,3]) 6 >>> sum(range(1,3)) 3 还有一个比较有意思的用法 a = range(1,11) b = range(1,10) c =  sum([item for item in a if item in b]) print c 输出: 45…

三者的作用: Number(): 可以用于任何数据类型转换成数值: parseInt().parseFloat(): 专门用于把字符串转换成数值: 一.Number( ): (1)如果是Boolean值,true和false将分别转换为1和0. (2)如果是数字值,只是简单的传入和返回. (3)如果是null值,返回0. (4)如果是undefined,返回NaN. (5)如果是字符串,遵循下列规则: 如果字符串截去开头和结尾的空白字符后,不是纯数字字符串,那么最终返回结果为NaN. 如果是字符…

Validate if a given string is numeric. Some examples:"0" => true" 0.1 " => true"abc" => false"1 a" => false"2e10" => true 解决方案: public class Solution { public boolean isNumber(String s) { in…

问题描述: Given a non-empty array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Example 1: Input: [2,2,1] Ou…

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return -1 instead. Have you met this question in a real interview?     Example Given the array [2,3,1,2,4,…

errors made, boundary conditions, <= vs < , decreasing vs increasing , ++, –, '0'/'1' vs 0/1 prototype of sum, return the starting position of c-style string containing the sum, just like sprintf return number of characters successfully read. p1=sum…

#include<stdio.h> int main(void) { intcount ,sum,aninterger; printf("enterthe interger anf terminute with negtive number\n"); count=0; sum=0; printf("enternumber %d:\n",count+1); scanf("%d",&aninterger); while(anint…

Write an algorithm to determine if a number is "happy". A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the…

+ 92 = 82 82 + 22 = 68 62 + 82 = 100 12 + 02 + 02 = 1 思路: (1)题意为判断给定的整数是否为一个"快乐的数",所谓快乐的数需要满足一下几个条件:将该整数的每个位上的数字的平方相加得到一个新的整数,循环对新的整数进行上述操作,如果最后所得整数收敛于1,则这样的数字为一个"快乐的数". (2)首先,判断0肯定不是一个"快乐的数":其次,对初始数字的每个位上数的平方相加,循环进行前面的操作:需要注…

136. Single Number 意思就是给你一堆数,每个数都出现了两次,只有一个数只出现了一次,找出这个数 位运算(和c艹一样) &:按位与 |:按位或 ^:异或(一样为0,不一样为1) 再说一下异或的性质,满足交换律和结合律 因此: 对于任意一个数n n ^ 0 = n n ^ n = 0 对于这道题来说,所有数依次异或剩下的就是那个数了 class Solution(object): def singleNumber(self, nums): ans = 0 for i in nums…