Find two numbers that appear once in the array!

Implementation code:

1, PHP implementation:

<?php     $arr =array (1,1,5,3,2,2);         // Because each element needs to be compared with other elements separately     //need a double loop     for ($i =0; $i <count ($arr); $i + + ) {                for ($j =0; $j <count ($arr), $j + +) {                     //The purpose here is to avoid the array elements and compare themselves              if ($i  !=  $j) {                                  //find the equivalent of two elements and set their values to null                  //If there is an indication that the value is null later, there is no need to assign the value to null         again.     &nbSp;   if ($arr [$i] ==  $arr [$j] &&  $arr [$j] != null] {                                       $arr [$i] = $arr [$j]=null;                                          //is here to avoid the meaningless comparisons of J                      break;                 }                              }        }        echo  $arr [$i]. ' <br/> ';    }>

2.js implementations:

for (var i in a) {for (Var j in a) {if (i! = j) {if (a[i] = = A[j] && a[j]! = ") {                A[i]=a[j]= ";        Break }}} Document.writeln (A[i]);}

This article is from the "Everything Possible" blog, please be sure to keep this source http://noican.blog.51cto.com/4081966/1597380

Find two numbers that appear once in the array!