java方法中的数组，执行什么操作？

Question

I have been struggling with this java problem for several days now, and now I have to give up. I have been told the answer which should be 5 5 3 3, but I cannot in any way see how it is possible to get that result.

Given the following java method:

public int[] methodName(int[] nums)
{
     int largestOdd=0;
     for(int i=nums.length-2;i>=0;i--)
     {
          if (nums[i+1] % 2 != 0 && nums[i+1] > largestOdd)
               largestOdd = nums[i+1];

          if (nums[i] == 0)
               nums[i] = largestOdd;
     }
     return(nums);
}

What is printed when the following Java statements are executed?

int[] nums = {0,5,0,3};
nums = methodName(nums)
for (int i = 0; i<nums.length;i++)
    System.out.print(nums[i] + "");
System.out.println();

It just doesnt make any sense for me that first of all it will start printing "5". In my opinion it should be "3" because nums[2+1] = 3 (last index element)

Second of all why will it print four numbers when the loop in the method only will loop through 3 times until hitting -1 ?

If someone can explain how to get the result in a understandable way, I would be very happy.

Thanks in advance

Answer 1

methodName runs backwards through the array, examining every pair of numbers. In this case, the first pair examined will be (0,3).

As it runs over the pairs, methodName keeps track of the largest odd number seen (it looks at the second number of each pair for this).

Whenever the first number is zero, it sets it to the largest odd number seen so far.

So in this case, it will:

1. Look at (0,3).
2. Is 3 odd? Yes. Is it the largest odd number seen so far? Yes. So keep track of 3.
3. Is 0 zero? Yes. So set it to 3. Now the array is {0, 5, 3, 3}.
4. Move our indices back by 1 and look at (5, 3).
5. Is 3 odd? Yes. Is it the largest odd number seen so far? No.
6. Is 5 zero? No.
7. Move our indices back by 1 and look at (0, 5).
8. Is 5 odd? Yes. Is it the largest odd number seen so far? Yes. So keep track of 5.
9. Is 0 zero? Yes. So set it to 5. Now the array is {5, 5, 3, 3}.
10. We are already at the beginning of the array, so we can't go back any further.
11. Return to the main method, and print the contents of the array.

Answer 2

{0,5,0,3}

for(int i=nums.length-2;i>=0;i--) // starts at 0 (index 2) <-> num.length 4 - 2

   // runs 3 times  num.length(4) - 2 = 0

   if (nums[i+1] % 2 != 0 && nums[i+1] > largestOdd)
           largestOdd = nums[i+1]; <-->  // 3

           // if odd and grater than 0 
           // first iteration largestOdd = 3


   if (nums[i] == 0)   // still i is index 2 {0, 5, 0, 3} = 0, so true
          nums[i] = largestOdd;  // nums[i] (index 2) = 3

          // after first iteration
          {0, 5, 3, 3}


  Second iteration do nothing (0 is not odd)

  Third iteration, do same as first iteration, making final array
  {5, 5, 3, 3}

Next Part

 // This method returns the same array you passed into to
 public int[] methodName(int[] nums)
 {
     return(nums);
  }

 // So
 nums = methodName(nums) = {5, 5, 3, 3} The new array produced by the method

Answer 3

you can analyse step by step.
In the for-loop: i starts from 2 to 0;
num = [0,5,0,3]
1.
    i = 2; largestOdd = 0;   nums[i+1] = nums[3] = 3;   nums[i] = nums[2] = 0;
    first condition "(nums[i+1] % 2 != 0 && nums[i+1] > largestOdd)" is true
        largestOdd = nums[i+1] = 3
    second condition (nums[i] == 0) is true
        nums[i] = nums[2] = largestOdd = 3
    nums = [0,5,3,3];
 2.
    i=1;   largestOdd = 3;   nums[i+1] = nums[2] = 3;   nums[i] = nums[1] = 5;
    first condition is false;
    second condition is false;
    nums = [0,5,3,3];
 3.
    i=0;  largestOdd = 3;   nums[i+1] = nums[1] = 5;   nums[i] = nums[0] = 0;
    first condition is true
         largestOdd = 5;
    second condition nums[0] = 0    is true
         nums[0] = largestOdd = 5;
    nums = [5,5,3,3];

Answer 4

Loop 1 start i=2,nums=[0,5,0,3],largestOdd=0
because:(nums[2+1]=3)/2 !=0 && (nums[2+1]=3)>(largestOdd=0)
so:largestOdd=(nums[2+1]=3)=3
beacuse:(nums[2]=0) ==0
so:nums[2]=(largestOdd=3)=3
Loop 1 end   i=2,nums=[0,5,3,3],largestOdd=3

Loop 2 start i=1,nums=[0,5,3,3],largestOdd=3
because:(nums[1+1]=3)/2 !=0 && (nums[1+1]=3)>(largestOdd=3)
so:next
beacuse:(nums[1]=5) !=0
so:next
Loop 2 end   i=1,nums=[0,5,3,3],largestOdd=3

Loop 3 start i=0,nums=[0,5,3,3],largestOdd=3
because:(nums[0+1]=5)/2 !=0 && (nums[0+1]=5)>(largestOdd=3)
so:largestOdd=(nums[0+1]=5)=5
beacuse:(nums[0]=0) ==0
so:nums[0]=(largestOdd=5)=5
Loop 3 end   i=0,nums=[5,5,3,3],largestOdd=5

END LOOP