Ruby's Louvre

每天学习一点点算法

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返回数组中不重复的元素

这是实现结构伪类only-of-type的部分代码：

    var ret= ["span","span","strong","span","b"]
    var norepeat = function(array){
      var set = array.join(",")+",";
      while(array.length){
        var el = array.shift();
        set = set.replace(el+",","");
        if(set.indexOf(el+",") != -1){
          set = set.replace(new RegExp(el+",","g"),"");
          array = set.replace(/,$/g,"").split(",");
        }else{
          set = el+","+set;
        }
      }
      return set.replace(/,$/g,"").split(",")
    }
    alert(norepeat(ret))//最后返回b,strong

觉得效果不高，向无忧高手求救。下面是客服果果的(好强大的正则替换啊)：

var ret= ["span","span","strong","span","b"];
var norepeat=function (x) {
  var y='c0cb5f0fcf239ab3d9c1fcd31fff1efc';
  return x.sort().join(y).replace(RegExp('(.*('+y+'|$))\\1+','g'),'').split(y);
};
alert(norepeat(ret));

abcdreamer 的实现（此实现能过滤非纯字符串数组里的重复，我比较喜欢）：

var a = ["span","span","strong","span","b"];
function norepeat(a) {
  var i = a.length, j, b;
  while (i--) {
    j = i;
    b = false;
    while (j--) {
      if (a[i] === a[j]) {
        a.splice(j, 1);
        i--;
        b = true;
      }
    }
    if (b) a.splice(i, 1);
  }
  return a;
}
alert(norepeat(a));

增加campaign 提供的解决方案！

	var ret= [{a:1},{a:1},"strong","span","span","b"],r=[];
	 for(var i=0,l=ret.length,tmp={};i<l;i++){
		tmp[ret[i]]?tmp[ret[i]]++:(tmp[ret[i]]=1);
	 }
	 for(var p in tmp){
		if(tmp[p]==1)
			r.push(p);
	 }
	 alert(r)

哪位高手有更好的实现，请告诉我一声！

看来有些人不明白norepeat函数与unique函数的区别，norepeat是去除数组中重复的元素，如果数组中有两个a，则去除这两个，unique则会去除其中之一。下面是unique函数的实现：

      Array.prototype.inArray = function(el){
        for (var i=0,n=this.length;i<n;i++ )
          if(this[i]===el)return true;
        return false;
      }
      Array.prototype.unique = function() {
        var i = 0, n = this.length,ret = [];
        for (; i < n; i++)
          if (!ret.inArray(this[i])) ret.push(this[i]);
        return ret;
      };
      var ret= ["span","span","strong","span","b"]
      alert(ret.unique())  //返回"span","strong","b"

posted on 2009-12-08 20:31  司徒正美  阅读(2978)  评论(11)  编辑  收藏  举报