判断数组中是否有重复元素,最容易想到的方法是使用2重循环,逐个遍历,比较,但是这个是最慢,最笨的方法,百度得出了更好的方法. var ary = new Array("111","22","33","111"); var nary=ary.sort(); for(var i=0;i<ary.length;i++){ if (nary[i]==nary[i+1]){ alert("数组重复内容:"+na
var arr=["a","b","c","d","c","b","d","f"];var newarr=[];//存放不重复元素的数组var count=[];//记录重复元素的元素和索引值for(var i=0;i<arr.length;i++){ var type=true; for(var j=0;j<newarr.length
验证JS中是否包含重复元素,有重复返回true:否则返回false 方案一. function isRepeat(data) { var hash = {}; for (var i in data) { if (hash[data[i]]) { return true; } // 不存在该元素,则赋值为true,可以赋任意值,相应的修改if判断条件即可 hash[data[i]] = true; } return false; } 方案二. function isRepeat(arrs) { i
方法一:正则 var ary = new Array("111","ff","222","aa","222"); alert(mm(ary)) // 验证重复元素,有重复返回true:否则返回false function mm(a) { return /(\x0f[^\x0f]+)\x0f[\s\S]*\1/.test("\x0f"+a.join("\x0f\x0f&qu
题目:比较传入函数的参数,将参数组成数组,从小到大排序,返回新的数组. 如: insert();console.log(arr); //[] insert(-1,-2); console.log(arr);//[-2,-1] insert(3);console.log(arr);//[-2,-1,3] insert(6,4,5);console.log(arr); //[-2,-1,3,4,5,6] 代码实现: var arr = []; var index = 0; function inse
其实蛮容易实现的,关键是简洁与否,下面是我自己写的. function randomSort(a){ var arr = a, random = [], len = arr.length; for (var i = 0; i < len; i++) { var index = Math.floor(Math.random()*(len - i)); random.push(a[index]); arr.splice(index,1); } return random; } var a = [1,
export function deteleObject(obj) { var uniques = []; var stringify = {}; for (var i = 0; i < obj.length; i++) { var keys = Object.keys(obj[i]); keys.sort(function(a, b) { return (Number(a) - Number(b)); }); var str = ''; for (var j = 0; j < keys.le
var data = [{}, {}, {}, {Id:1}] var datawilldele = [];//2,4,5 data.forEach(function (v, i,arry) { if ($.isEmptyObject(v)) { datawilldele.push(i); } }); for (var i = 0; i < datawilldele.length; i++) { data.splice(datawilldele[i]-i,1); }
/* 调用完该方法,原数组只留下非重复的数据 返回一个数组,里面是依次出现的重复元素 */Array.prototype.distinct = function () { var removeArr = [], retainArr = []; for (var i = 0; i < this.length; i ++) { var elem = this[i]; if (this.indexOf(elem) !== i ) { rem