Java，如果数组中“ x”在“ y”旁边，则返回true

Question

This is, I imagine, a really simple problem to solve however I just can't figure it out.

An array contains a list of integers and I want to return true if every number 'x' in the array is followed by the number 'y'.

So arrays with {x,3,4,y} or {x,x,y,4,5} or {5,8,x,x} would be false .

Whereas arrays with {x,y,4,1} or {x,y,5,1,x,y} would be true .

This is what I have tried so far:

for (int i = 0; i < nums.length-1; i++)
{
    if (nums[i] == x && nums[i+1] == y)
    {
        return true;
    }
    else
    {
        return false;
    }
}

return false;

My code however will only work for the first two elements in the array (so 0 and 1). It won't detect any integers further down in the array, so how do I do this?

Thank you.

Answer 1

I want to return true if every number 'x' in the array is followed by the number 'y'.

You need to get rid of the else and modify the checks like so:

for (int i = 0; i < nums.length - 1; i++)
{
    if (nums[i] == x && nums[i + 1] != y)
    {
        return false;
    }
}

return true;

Caveats:

1. This returns true if x is not present in the array. It's unclear from the question whether this is the behaviour you want.
2. This does not check whether x is the last element of the array. Again, it's not entirely clear what you'd expect to happen if it is.

Answer 2

for (int i = 0; i < nums.length-1; i++) {    
    if (nums[i] == 2 && nums[i+1] != 3) {
        return false;
    }
}

return true;

Answer 3

This code will return False iff the 'x' number is not followed by 'y' .

for (int i = 0; i < nums.length-1; i++)
{
    if(nums[i] == 2 && nums[i+1] != 3) {
        return false;
    }
}
return true;

Answer 4

When the code reaches a return statement it stops executing the function and returns the specified value, that's why your for is only executed once.

Since you want the condition to apply to ALL the elements in the array, you have to return false when you find a 2 that isn't followed by a 3. If the 2 is followed by a 3, you just keep checking the next position.

Like this:

for (int i = 0; i < nums.length-1; i++)
{ 
     if(nums[i] == 2 && nums[i+1] != 3)
     {
          return false;
     }
}
return true;

Answer 5

for (int i = 0; i < nums.length-1; i++)
  {
    if(nums[i] == 2 && nums[i+1] == 3)
    {
    return true;
    }

  }
  return false;

Answer 6

Your Code is returning in the first iteration always. because you have return written there in both If and else block.

remove the else case or just the return; from else case if there is any other logic inside.

int x = 2;
int y=3;
    for (int i = 0; i < nums.length-1; i++) {    
        if (nums[i] == x && nums[i+1] != y) {
            return false;
        }
    }
return true;

If you do not have x in the array, it will return true. and if you have any x that does not have y after it, it will return false.

Answer 7

Try out Following

    boolean flag = false;
      for (int i = 0; i <= (nums.length-1); i++) {
         if(nums[i] == x) {
              if (i == (nums.length-1)) {
                  if(nums[i] == x) {
                      flag = false;
                  }
              } else {
                  if(nums[i+1] == y) {
                      flag = true;
                  } else {
                      flag = false;
                      break;
                  }
              }
           }
      }
      return flag;

Answer 8

You can try out the following:

if(nums[nums.length - 1] == x)
  {
  return false;
  }
else
  {
  boolean flag = false;

  for(i = 0; i < nums.length - 1; i++)
    {
    if(nums[i] == x)
      {
      if(nums[i + 1] == y)
        {
        flag = true;
        }
      else
        {
        flag = false;
        break;
        }
      }
    }

  return flag;
  }