Codewars裏的 6kyu Kata。
題目說明：
You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer
N. Write a method that takes the array as an argument and returns this "outlier"N.Examples
[2, 4, 0, 100, 4, 11, 2602, 36] Should return: 11 (the only odd number) [160, 3, 1719, 19, 11, 13, -21] Should return: 160 (the only even number)
解題代碼：
public class FindOutlier{
static int find(int[] integers){
int i = 0, j = integers.length - 1;
if(Math.abs(integers[i])%2 != Math.abs(integers[j])%2) {// 不同
i++;
if(Math.abs(integers[i])%2 == Math.abs(integers[j])%2) {// 進一步判斷相同
return integers[i-1];
}
j--;
if(Math.abs(integers[i])%2 == Math.abs(integers[j])%2) {
return integers[j+1];
}
}
while(i < j) {
if(Math.abs(integers[i])%2 != Math.abs(integers[j])%2) {
if(Math.abs(integers[i])%2 == Math.abs(integers[0])%2) {
return integers[j];
}else {
return integers[i];
}
}
i++;
j--;
}
return integers[i];
}
}
Test Cases:
import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;
import java.util.Random;
public class OutlierTest {
@Test
public void test() {
/* Basic tests */
int[] ints1 = { 2, 6, 8, 10, 3 }; // odd at the back
int[] ints2 = { 2, 6, 8, 200, 700, 1, 84, 10, 4 }; // odd in the middle
int[] ints3 = { 17, 6, 8, 10, 6, 12, 24, 36 }; // odd in the front
int[] ints4 = { 2, 1, 7, 17, 19, 211, 7 }; // even in the front
int[] ints5 = { 1, 1, 1, 1, 1, 44, 7, 7, 7, 7, 7, 7, 7, 7 }; // even in the middle
int[] ints6 = { 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 35, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 1000 }; // even
// at
// the
// end
int[] ints7 = { 2, -6, 8, -10, -3 }; // odd at the back, negative
int[] ints8 = { 2, 6, 8, 2, -66, 34, -35, 66, 700, 1002, -84, 10, 4 }; // odd in the middle, negative
int[] ints9 = { -1 * Integer.MAX_VALUE, -18, 6, -8, -10, 6, 12, -24, 36 }; // odd in the front, negative
int[] ints10 = { -20, 1, 7, 17, 19, 211, 7 }; // even in the front, negative
int[] ints11 = { 1, 1, -1, 1, 1, -44, 7, 7, 7, 7, 7, 7, 7, 7 }; // even in the middle, negative
int[] ints12 = { 1, 0, 0 }; // odd answer, zeroes
int[] ints13 = { 3, 7, -99, 81, 90211, 0, 7 }; // even in the middle, zero
assertEquals(3, FindOutlier.find(ints1));
assertEquals(1, FindOutlier.find(ints2));
assertEquals(17, FindOutlier.find(ints3));
assertEquals(2, FindOutlier.find(ints4));
assertEquals(44, FindOutlier.find(ints5));
assertEquals(1000, FindOutlier.find(ints6));
assertEquals(-3, FindOutlier.find(ints7));
assertEquals(-35, FindOutlier.find(ints8));
assertEquals(-1 * Integer.MAX_VALUE, FindOutlier.find(ints9));
assertEquals(-20, FindOutlier.find(ints10));
assertEquals(-44, FindOutlier.find(ints11));
assertEquals(1, FindOutlier.find(ints12));
assertEquals(0, FindOutlier.find(ints13));
// Random tests
Random r = new Random();
int positionA = r.nextInt(400);
int positionB = r.nextInt(400);
int[] randomOdds = new int[400];
int[] randomEvens = new int[400];
int evenSample = r.nextInt(3000) * 2;
int oddSample = r.nextInt(3000) * 2 + 1;
for (int i = 0; i < 400; i++) {
randomOdds[i] = r.nextInt(3000) * 2 + 1;
}
for (int i = 0; i < 400; i++) {
randomEvens[i] = r.nextInt(3000) * 2;
}
randomOdds[positionA] = evenSample;
randomEvens[positionB] = oddSample;
assertEquals(evenSample, FindOutlier.find(randomOdds));
assertEquals(oddSample, FindOutlier.find(randomEvens));
// large array test
int positionC = r.nextInt(10000000);
int positionD = r.nextInt(10000000);
int[] bigOddArray = new int[10000000];
int[] bigEvenArray = new int[10000000];
for (int i = 0; i < 10000000; i++) {
bigOddArray[i] = r.nextInt(100000) * 2 + 1;
bigEvenArray[i] = r.nextInt(100000) * 2;
}
bigOddArray[positionC] = evenSample;
bigEvenArray[positionD] = oddSample;
assertEquals(evenSample, FindOutlier.find(bigOddArray));
assertEquals(oddSample, FindOutlier.find(bigEvenArray));
}
}
個人總結：
import java.util.Arrays;
public class FindOutlier {
public static int find(int[] integers) {
// Since we are warned the array may be very large, we should avoid counting
// values any more than we need to.
// We only need the first 3 integers to determine whether we are chasing odds or
// evens.
// So, take the first 3 integers and compute the value of Math.abs(i) % 2 on
// each of them.
// It will be 0 for even numbers and 1 for odd numbers.
// Now, add them. If sum is 0 or 1, then we are chasing odds. If sum is 2 or 3,
// then we are chasing evens.
int sum = Arrays.stream(integers).limit(3).map(i -> Math.abs(i) % 2).sum();
int mod = (sum == 0 || sum == 1) ? 1 : 0;
return Arrays.stream(integers).parallel() // call parallel to get as much bang for the buck on a "large" array
.filter(n -> Math.abs(n) % 2 == mod).findFirst().getAsInt();
}
}
public class FindOutlier {
static int find(int[] integers) {
int oddcount = 0, odd = 0, evencount = 0, even = 0;
for (int i : integers) {
if (i % 2 == 0) {
even = i;
evencount++;
} else {
odd = i;
oddcount++;
}
if (evencount > 0 && oddcount > 0) {
if (evencount > 1) {
return odd;
} else if (oddcount > 1) {
return even;
}
}
}
return evencount > 1 ? odd : even;
}
}
import java.util.Arrays;
public class FindOutlier {
static int find(int[] integers) {
int[] array = Arrays.stream(integers).filter(i -> i % 2 == 0).toArray();
return array.length == 1 ? array[0] : Arrays.stream(integers).filter(i -> i % 2 != 0).findAny().getAsInt();
}
}
public class FindOutlier {
static int find(int[] integers) {
int even = 0;
int odd = 0;
int cycle = 0;
for (Integer value : integers) {
if (value % 2 == 0) {
cycle++;
even = value;
} else {
odd = value;
}
}
return (cycle > 1) ? odd : even;
}
}