#include<iostream> #include<ctime> #include<random> using namespace std; void knuth(int n, int m) { srand((unsigned int)time(NULL)); ; i < n; i++) { /* 注意到在整个for循环中,随机数种子已经固定,rand() 的值是不变的 这里n必须减去i,否则很有可能产生的随机数量小于n */ if (rand() % (n
我们用SQL查询数据时后,基于某些原因不想看到某字段的值,比如密码,我们可以通过创建视图,忽略某一字段的值. 同时我们也可以直接通过SQL语句来让其显示某个固定值: (1)一般查询语句: SELECT id,userName,PASSWORD FROM t_user; (2)让某字段查询结构显示为固定值: SELECT id,userName,'*****' PASSWORD FROM t_user; 或: SELECT id,userName,'*****' AS PASSWORD FROM
算法导论:22页2.3-7 描述一个运行时间为O(nlogn)的算法,找出n个元素的S数组中是否存在两个元素相加等于给定x值 AC解: a=[1,3,6,7,9,15,29] def find2sumx(nums,x): nums.sort() le,ri=0,len(nums)-1 while le>=0 and ri<=len(nums) and le<ri: if nums[le]+nums[ri]<x: le+=1 elif nums[le]+nums[ri]>x:
最近被算法虐了一下,刷一下leetcode,找找存在感 如题: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Exam
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m
package com.java.tencent; import java.lang.reflect.Array; import java.util.Arrays; import java.util.HashMap; import java.util.Map; public class T_1_twoSum { /*初级解法*/ public int[] twoSum(int[] nums, int target) { int[] result = new int[2]; for(int i=0
