Problem:
Given an int array, return a new array with double the length where its last element is the same as the original array, and all the other elements are 0. The original array will be length 1 or more. Note: by default, a new int array contains all 0's.
makeLast({4, 5, 6}) → {0, 0, 0, 0, 0, 6}
makeLast({1, 2}) → {0, 0, 0, 2}
makeLast({3}) → {0, 3}
Solution:
public int[] makeLast(int[] nums) {
int len = nums.length;
int[] myArray = new int[2*len];
myArray[myArray.length-1] = nums[len-1];
return myArray;
}
public int[] makeLast(int[] nums) {
ReplyDeleteint len = nums.length;
int[] bs = new int[2*len];
bs[bs.length-1] = nums[len-1];
return bs;
}
private final int[] makeLast(int[] nums) {
ReplyDeletefinal int[] intArray = new int[nums.length * 2];
intArray[intArray.length - 1] = nums[nums.length - 1];
return intArray;
}
public int[] makeLast(int[] nums) {
ReplyDeleteint[]a=new int[2*nums.length];
a[a.length-1]=nums[nums.length-1];
return a;
}