判斷數值數組中各個數字出現的奇偶次數
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>数组操作</title>
</head>
<body>
<script type="text/javascript">
var arr=[3,1,2,2,1,3,1];
var sum=[];
var res=[];
var count=0;
var temp;
for(var i=0;i<arr.length;i++){
if(res.indexOf(arr[i])==-1){
res.push(arr[i]);
}
}
for(var i=0;i<res.length;i++){
for(var j=0;j<arr.length;j++){
if(arr[j]==res[i]){
count++;
}
}
sum.push(count);
count=0;
}
console.log(res);//[3,1,2]
for(var i=0;i<res.length;i++){
var str=(sum[i]%2==0)?"偶数":"奇数";
console.log(res[i]+"出现了"+sum[i]+"次");
console.log(res[i]+"出现了"+str+"次");
}
</script>
</body>
</html>阿里筆試-數組操作-找出兩個數組中不同的元素
<script type="text/javascript">
function diff(arr1,arr2){
var ress = [];
var arr = arr1.concat(arr2);
for(var i=0,len=arr.length;i<len;i++){
if((arr1.indexOf(arr[i])>=0 && arr2.indexOf(arr[i])<0) || (arr1.indexOf(arr[i])<0 && arr2.indexOf(arr[i])>=0)){
ress.push(arr[i]);
}
}
return ress;
}
var arr1 = [1,2,3,5,7,6];
var arr2 = [1,2,5];
var res = diff(arr1,arr2);
console.log(res);//[3, 7, 6]
</script>數組去重
方法1
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>数组去重01</title>
</head>
<body>
<script type="text/javascript">
//给数组原型添加方法
Array.prototype.unique = function(){
var arr = [];
for(var i=0,i=this.length;i<len;i++){
if(arr.indexOf(this[i]) == -1){
arr.push(this[i]);
}
}
return arr;
};
console.log([1,2,3,2,5,6,3].unique());//[1, 2, 3, 5, 6]
</script>
</body>
</html><!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>数组去重02</title>
</head>
<body>
<script type="text/javascript">
Array.prototype.unique = function(){
var n = {},
r=[]; //n为哈希表，r为临时数组
for(var i = 0; i < this.length; i++) //遍历当前数组
{
if (!n[this[i]]) //如果hash表中没有当前项
{
n[this[i]] = true; //存入哈希表
r.push(this[i]); //把当前数组的当前项push到临时数组里面
}
}
return r;
}
console.log([1,2,3,2,5,6,3].unique());//[1, 2, 3, 5, 6]
</script>
</body>
</html>方法4
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>数组去重</title>
</head>
<body>
<script type="text/javascript">
Array.prototype.unique = function(){
var arr = [this[0]];//结果数组
for(var i=1;i<this.length;i++){//从第二项开始遍历
if(this.indexOf(this[i]) == i){
//如果当前数组的第i项在当前数组中第一次出现的位置不是i，那么表示第i项是重复的，忽略掉。否则存入结果数组
arr.push(this[i]);
}
}
return arr;
}
console.log([1,2,3,4,2,3,4].unique());// [1, 2, 3, 4]
</script>
</body>
</html>方法5
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<title>filter</title>
<link rel="stylesheet" href="">
</head>
<body>
<script type="text/javascript">
var arr = [4,5,3,2,3,4,5,1];
function fn(num){
var res = num.filter(function(item,index,array){
return num.indexOf(item) === index;//num.indexOf(item)将会返回元素在数组第一次出现的位置
//对于多次出现的元素，除第一次外，其他情况都返回false
});
return res;
}
console.log(fn(arr));//[4, 5, 3, 2, 1]
</script>
</body>
</html>
