Question

Given an array of ints, return true if 6 appears as either the first or last element in the array. The array will be length 1 or more.

Answer 1

// Given an array of ints, return true if 6 appears as either the first or last element in the array.

// The array will be length 1 or more.

public boolean firstLast6(int[] nums)

{ return (nums[0] == 6 || nums[nums.length-1] == 6); }

// Given an array of ints, return true if the array is length 1 or more, and the

// first element and the last element are the same.

public boolean sameFirstLast(int[] nums)

{ return (nums.length >= 1 && nums[0] == nums[nums.length-1]);}

// Return an int array length 3 containing the first 3 digits of pi, {3, 1, 4}.

public int[] makePi()

{

int[] pi = {3, 1, 4};

return pi;

}

// Given 2 arrays of ints, a and b, return true if they have the same first element or they have the same last element.

// Both arrays will be length 1 or more.

public boolean commonEnd(int[] a, int[] b)

{ return (a[0] == b[0] || a[a.length-1] == b[b.length-1]); }

// Given an array of ints length 3, return the sum of all the elements.

public int sum3(int[] nums)

{ return (nums[0] + nums[1] + nums[2]); }

// Given an array of ints length 3, return an array with the elements

// "rotated left" so {1, 2, 3} yields {2, 3, 1}.

public int[] rotateLeft3(int[] nums)

{

int[] rotated = {nums[1], nums[2], nums[0]};

return rotated;

}

// Given an array of ints length 3, return a new array with the elements in reverse order,

// so {1, 2, 3} becomes {3, 2, 1}.

public int[] reverse3(int[] nums)

{

int[] reversed = {nums[2], nums[1], nums[0]};

return reversed;

}

// Given an array of ints length 3, figure out which is larger between the first and last elements

// in the array, and set all the other elements to be that value. Return the changed array.

public int[] maxEnd3(int[] nums)

{

int[] maxVal = new int[3];

maxVal[0] = nums[0];

if(nums[2] >= maxVal[0])

maxVal[0] = nums[2];

maxVal[1] = maxVal[0];

maxVal[2] = maxVal[0];

return maxVal;

}

// Given an array of ints, return the sum of the first 2 elements in the array.

// If the array length is less than 2, just sum up the elements that exist,

// returning 0 if the array is length 0.

public int sum2(int[] nums)

{

if(nums.length >= 2)

return (nums[0] + nums[1]);

if(nums.length == 1)

return nums[0];

return 0;

}

// Given 2 int arrays, a and b, each length 3, return a new array length 2

// containing their middle elements.

public int[] middleWay(int[] a, int[] b)

{

int[] mids = {a[1], b[1]};

return mids;

}

// Given an array of ints, return a new array length 2 containing the first and last

// elements from the original array. The original array will be length 1 or more.

public int[] makeEnds(int[] nums)

{

int[] temp = {nums[0], nums[nums.length-1]};

return temp;

}

// Given an int array length 2, return true if it contains a 2 or a 3.

public boolean has23(int[] nums)

{

if(nums[0] == 2 || nums[0] == 3)

return true;

return (nums[1] == 2 || nums[1] == 3);

}

// Given an int array length 2, return true if it does not contain a 2 or 3.

public boolean no23(int[] nums)

{

if(nums[0] == 2 || nums[0]

Answer 2

Answer:

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