比如这个数组 var a=[["1","3","10"],["3","1","10"],["9","2"],["1","10","3"]];我想去掉["1","3","10"],["3","1","10"]["1","10","3"] 这3个元素,有什么好办法吗

解决方案 »

  1.   

    我就这办法比如这个["1","3","10"]拿出1与之后的比较比较出都含有1出return["3","1","10"],["1","10","3"]然后再用第一个的3比较,看挑出来的两个是否还有3没想到什么好方法
      

  2.   

    先判断元素个数拿3个元素比较xy与xz与xy比较全部相等就说明三个元素相同如何?
      

  3.   

    xy与xz与yz比较两两求积,如何?
      

  4.   

    抛砖引玉var a=[["1","3","10"],["3","1","10"],["9","2"],["1","10","3"]];var b = [];
    for(var i in a) b.push(a[i].sort().join(','));
    b.sort();
    var c = [];
    for(i=1; i<b.length; i++) {
      if(b[i] != b[i-1]) c.push(b[i]);
    }
    for(i in c) c[i] = c[i].split(',');
    alert(c);
      

  5.   

    看到前面的玉,也写一个自己的
    var a=[["1","3","10"],["3","1","10"],["9","2"],["1","10","3"]];var ta=a.join(",").split(",").sort();//转换成为一维数组进行排序var i = 0;var b = [];function number()
    {
    for(i=0; i<ta.length-1; i++)
    {
    if(ta[i] != ta[i+1])
    {
    b.push(ta[i+1])
    }
    };
    b.splice(0,1);
    b.splice(1,1);
    document.write(b);
    };number();