JavaScript获取多个数组的交集简单实例_javascript技巧

在项目中需要取得多个数组的交集，故本例只是适用于特定场景。比如A数组 var a = {1000,10001,10002,10003}; B数组 var b = {10002, 10003}; C数组var c = {10003}; 需要取得这三个数组的交集数组。
      具体思路为：先初始一个最小数组为A数组,元素个数最小的也是数组A的长度.然后再迭代各个数组,取得这几个数组长度最短的为最小长度,同时得到长度最短数组。然后,迭代最小数组与各个数组,开始比较元素相等,使用计数器来判断是否该元素在各个数组中都存在即为交集元素.
      思路比较简单,但是可以实现多数组取得交集,代码如下：

复制代码 代码如下:

 
 
 
 
 
 
  <script> <BR> function getValues(obj){ <BR> var values = ""; <BR> var l = obj.options.length; <BR> for (var i=0; i<l; i++) { <BR> if (i != (l-1)) { <BR> values += obj.options(i).value + "_"; <BR> } <BR> else { <BR> values += obj.options(i).value; <BR> } <BR> } <BR> return values; <BR> } <br><br> function _test() { <BR> var ids = getValues(document.all.aa); <BR> var aa = _getIntersection(ids); <BR> } <br><br> function _getIntersection(src) { <BR> var tAry = src.split("_"); <BR> //最小数组 <BR> var minAry = null; <BR> var min = tAry[0].split(",").length; //初始化第一个为长度最小的数组 <BR> minAry = tAry[0].split(","); <BR> for (var i = 1, len = tAry.length; i<len; i++) { <BR> var temp = tAry[i].split(","); <BR> if (temp.length < min) { <BR> min = temp.length; <BR> minAry = temp; <BR> } <BR> } <BR> alert("最小数组："+minAry); <br><br> var ret = ''; <BR> for (var i = 0, len = minAry.length; i<len; i++) { <BR> var srcNum = parseInt(minAry[i]); <BR> var counter = 0; <BR> for (var j = 0, ll = tAry.length; j<ll; j++) { <BR> var tt = tAry[j].split(","); <BR> for (var k = 0, l = tt.length; k<l; k++) { <BR> var tarNum = parseInt(tt[k]); <BR> if (srcNum == tarNum) { <BR> counter ++; <BR> } <BR> } <BR> } <BR> if (counter == tAry.length) { <BR> ret += srcNum + ","; <BR> } <BR> } <BR> ret = strSlice(ret, ','); <BR> alert("交集是：" + ret); <BR> } <br><br> //去掉结尾分隔符 <BR> function strSlice(str, split){ <BR> if ((str!=null && str!="") && (split!=' ')) <BR> return ((str.charAt(str.length-1) == split) ? str.substring(0, str.length-1) : str); <BR> else <BR> return str; <BR> } <BR> </script>

 

 
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