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
1.如何判断数组元素是否存在重复项 1)定义测试数组 //定义测试的数组(1个没有重复元素,1个有重复元素) var arr1 = new Array("111","333","222","444"); var arr2 = new Array("aa","cc","bb","dd","bb"); 2) 判断数组元素重复的方法
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
Maximum repetition substring Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8669 Accepted: 2637 Description The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same conse
//判断数组array中是否包含元素obj的函数,包含则返回true,不包含则返回false function array_contain(array, obj){ for (var i = 0; i < array.length; i++){ if (array[i] == obj)//如果要求数据类型也一致,这里可使用恒等号=== return true; } return false; }