Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k. 这道题跟之前两道Contains Duplicate 包含重复值和Conta
Given an array of integers and an integer k, return true if and only if there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k. (Old Version) Given an array of integers and an i
Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 这道题不算难题,就是使用一个哈希表,遍历整个数组,如果哈希表里存在,返回fal
public class Demo { /** * 去掉重复值 */ public static void main(String[] args) { String test = "100,120,166,1555,120,150,100"; String[] test1 = test.split(","); ArrayList list = new ArrayList(); for (int i = 0; i < test1.length; i++) { i
今天遇到了一个问题,就是从数据库中去除的数组为一个二维数组,现在就是想将二位数组进行去重,但是在php中,对于一个一维数组,我们可以直接使用php的系统函数array_unique,但是这个函数不能对多维数组进行去除重复,因此我需要自己写一个去除二维数组重复值的函数. function array_unique_fb($array2D){ foreach ($array2D as $v){ $v=join(',',$v);//降维,也可以用implode,将一维数组转换为用逗号连接的字符串 $t
主要思想: 数组可以无序 假设数字里的值都为正 循环判断数组 如果与前面的数字相同则变为-1 然后记录-1的个数算出重复值 然后重新new一个减去重复值长度的新数组 和原数组判断 不为-1的全部复制进来即可 代码如下: package Del_Same_Num; public class Del_Same_Num { static int count=0; //计算重复值 public static int count_same_number(int[] a) { for(int i=0;i<a
在生产环境中,我们有的列是不允许出现重复值的,亦或是某两列不允许同时重复,但由于前端未做限制,或者没限制住,出现了单列重复值,或者两列本应组成唯一组合却也出现重复,这两种情况都是不允许的.现在由于前端应用限制不住,要做删除操作后,添加唯一索引,从数据库层面进行限制,以下是处理过程: mysql> select * from aixuan1; +----+------+-------+ | id | text | text1 | +----+------+-------+ | 1 | aa