I am trying to compare elements of the same array. That means that i want to compare the 0 element with every other element, the 1 element with every other element and so on. The problem is that it is not working as intended. . What i do is I have two for loops that go from 0 to array.length-1.. Then i have an if statement that goes as follows: if(a[i]!=a[j+1])
for (int i = 0; i < a.length - 1; i++) {
for (int k = 0; k < a.length - 1; k++) {
if (a[i] != a[k + 1]) {
System.out.println(a[i] + " not the same with " + a[k + 1] + "\n");
}
}
}
First things first, you need to loop to < a.length rather than a.length - 1. As this is strictly less than you need to include the upper bound.
So, to check all pairs of elements you can do:
for (int i = 0; i < a.length; i++) {
for (int k = 0; k < a.length; k++) {
if (a[i] != a[k]) {
//do stuff
}
}
}
But this will compare, for example a[2] to a[3] and then a[3] to a[2]. Given that you are checking != this seems wasteful.
A better approach would be to compare each element i to the rest of the array:
for (int i = 0; i < a.length; i++) {
for (int k = i + 1; k < a.length; k++) {
if (a[i] != a[k]) {
//do stuff
}
}
}
So if you have the indices [1...5] the comparison would go
1 -> 2 1 -> 3 1 -> 4 1 -> 5 2 -> 3 2 -> 4 2 -> 5 3 -> 4 3 -> 54 -> 5So you see pairs aren't repeated. Think of a circle of people all needing to shake hands with each other.
Try this or purpose will solve with lesser no of steps
for (int i = 0; i < a.length; i++)
{
for (int k = i+1; k < a.length; k++)
{
if (a[i] != a[k])
{
System.out.println(a[i]+"not the same with"+a[k]+"\n");
}
}
}
for (int i = 0; i < a.length; i++) {
for (int k = 0; k < a.length; k++) {
if (a[i] != a[k]) {
System.out.println(a[i] + " not the same with " + a[k + 1] + "\n");
}
}
}
You can start from k=1 & keep "a.length-1" in outer for loop, in order to reduce two comparisions,but that doesnt make any significant difference.
Source: Stackoverflow.com