一 ,var dataArr = []; var data = new Date(); var year = data.getFullYear(); data.setMonth(data.getMonth()+1, 1)//获取到当前月份,设置月份 for (var i = 0; i < 12; i++) { data.setMonth(data.getMonth() - 1);//每次循环一次 月份值减1 var m = data.getMonth() + 1; m = m < 10 ? &
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
题目: Given an 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? 中文(给定一个整数的数组,每个整数都出现了两次,只有一个出现了一次,找到它 建
要求的格式 代码块 $('.btn-confirm').on('tap',function(){ var arr={}; var name = $("input[name='insurance_attributes[ty_beibaoren][ty_beibaoren_name]']").val(); var num = $("input[name='insurance_attributes[ty_beibaoren][ty_beibaoren_id_number]]']&q
JS: var arrData = [1,3,5,7,7,8,9,3,10,8,"sdsdsds","sss","ffff","sss","sss"];function uniqueArray(data){ data = data || []; var a = {}; for (var i=0; i<data.length; i++) { var v = data[i]; if (typeof(a[v