Back to the simple coding questions. This time with javascript – it seems the easiest language to test with.
Code for the problem can be run over here: http://js.do/code/svetliy-second-largest
var first_largest;
var second_largess;
function large(a_list){
if (a_list.length == 0 || a_list.length == 1)
return "None";
if (a_list[0] > a_list[1]){
first_largest = a_list[0];
second_largest = a_list[1];
} else {
first_largest = a_list[1];
second_largest = a_list[0];
}
for (var i=2; i<a_list.length; i++){
if (a_list[i] > first_largest){
second_largest = first_largest;
first_largest = a_list[i];
continue;
}
if (a_list[i] > second_largest){
second_largest = a_list[i];
}
}
return(second_largest);
}
var a_list = [1, 3, 4, 5, 0, 2];
document.write(large(a_list));