我正在嘗試編寫代碼來顯示偶數元素到偶數索引和奇數到奇數索引，如果添加的數字相同，則相應地添加零。例子：x = [1,2,3,4] 輸出：2 1 4 3 x = [1 1 1 4] 輸出：4 1 0 1 0 1我達到了偶數和奇數位置，但在那之后就卡住了。下面是我的代碼。import java.util.*;class ArrayDemo3 { public static void main(String[] args) { Scanner s = new Scanner(System.in); System.out.println("Enter Size of Array :: "); int size = s.nextInt(); int[] x = new int[size]; System.out.println("Array Created having the size :: " + size); System.out.println("Enter Elements for Array :: "); for (int i = 0; i < size; i++) { System.out.println("Enter element no-" + (i + 1) + " ::"); x[i] = s.nextInt(); } System.out.println("Contents of Array ::"); for (int i = 0; i < size; i++) { System.out.print(x[i] + " "); } for (int i = 0; i < size; i = i + 1) { int even = 0; int odd = 1; if (i < size && x[i] % 2 == 0) { System.out.print("even : "); even = even + i; System.out.print("position" + i + " " + x[i] + " "); } else { System.out.print("odd : "); odd = odd + i; System.out.print(i + " " + x[i] + " "); } if (even < size && odd < size) { int temp = x[even]; x[even] = x[odd]; x[odd] = temp; } else { } //System.out.print(x[i] + " "); } }}
3 回答
江戶川亂折騰
TA貢獻1851條經驗 獲得超5個贊
您可以將問題分為三部分：
首先創建兩個列表，一個包含按遇到的順序排列的偶數，另一個包含奇數：
private static List<List<Integer>> createOddityLists(int... numbers) {
List<Integer> numsList = Arrays.stream(numbers).boxed().collect(Collectors.toList());
List<List<Integer>> numsByOddity = new ArrayList<List<Integer>>();
numsByOddity.add(new ArrayList<>()); // List of odd numbers
numsByOddity.add(new ArrayList<>()); // List of even numbers
numsList.forEach(num -> numsByOddity.get(num % 2).add(num));
return numsByOddity;
}
用零 ( s) 填充兩個列表中較短的一個，0使其與另一個列表的長度相等：
private static void padShorterList(List<List<Integer>> numsByOddity) {
int sizeDiff = numsByOddity.get(0).size() - numsByOddity.get(1).size();
int listIndexToBePadded = sizeDiff < 0 ? 0 : 1;
List<Integer> padding = Collections.nCopies(Math.abs(sizeDiff), 0);
numsByOddity.get(listIndexToBePadded).addAll(padding);
}
最后連接兩個列表：
private static List<Integer> joinLists(List<List<Integer>> numsByOddity) {
List<Integer> resultList = new ArrayList<>(numsByOddity.get(1));
for (int idx = 0; idx < numsByOddity.get(0).size(); idx++)
resultList.add(idx * 2, numsByOddity.get(0).get(idx));
return resultList;
}
以下是完整的工作示例：
public class ArrayRearrangement {
public static void main(String[] args) {
// int[] result = rearrange(1, 2, 3, 4);
int[] result = rearrange(1, 1, 1, 4);
System.out.println(Arrays.stream(result).boxed().collect(Collectors.toList()));
}
private static int[] rearrange(int... numbers) {
List<List<Integer>> numsByOddity = createOddityLists(numbers);
padShorterList(numsByOddity);
return joinLists(numsByOddity).stream().mapToInt(i->i).toArray();
}
private static List<List<Integer>> createOddityLists(int... numbers) {
List<Integer> numsList = Arrays.stream(numbers).boxed().collect(Collectors.toList());
List<List<Integer>> numsByOddity = new ArrayList<List<Integer>>();
numsByOddity.add(new ArrayList<>()); // List of odd numbers
numsByOddity.add(new ArrayList<>()); // List of even numbers
numsList.forEach(num -> numsByOddity.get(num % 2).add(num));
return numsByOddity;
}
private static void padShorterList(List<List<Integer>> numsByOddity) {
int sizeDiff = numsByOddity.get(0).size() - numsByOddity.get(1).size();
int listIndexToBePadded = sizeDiff < 0 ? 0 : 1;
List<Integer> padding = Collections.nCopies(Math.abs(sizeDiff), 0);
numsByOddity.get(listIndexToBePadded).addAll(padding);
}
private static List<Integer> joinLists(List<List<Integer>> numsByOddity) {
List<Integer> resultList = new ArrayList<>(numsByOddity.get(1));
for (int idx = 0; idx < numsByOddity.get(0).size(); idx++)
resultList.add(idx * 2, numsByOddity.get(0).get(idx));
return resultList;
}
}
希望這可以幫助。
收到一只叮咚
TA貢獻1821條經驗 獲得超5個贊
根據索引對元素進行排序，即如果元素是偶數，則它必須位于偶數位置，反之亦然
int sortArrayByEvenOddIndex(int arr[]) {
int n = arr.length;
int res[] = new int[n];
int odd = 1;
int even = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 0) {
res[even] = arr[i];
even += 2;
} else {
res[odd] = arr[i];
odd += 2;
}
}
return res;
}
長風秋雁
TA貢獻1757條經驗 獲得超7個贊
使用數組我們可以做到這一點。代碼需要優化。
public static int[] arrangeInEvenOddOrder(int[] arr)
{
// Create odd and even arrays
int[] oddArr = new int[arr.length];
int[] evenArr = new int[arr.length];
int oCount = 0, eCount = 0;
// populate arrays even and odd
for (int i = 0; i < arr.length; i++) {
if (arr[i] % 2 == 0)
evenArr[eCount++] = arr[i];
else
oddArr[oCount++] = arr[i];
}
int[] resArr = new int[oCount >= eCount?
2*oCount : 2*eCount-1];
// populate elements upto min of the
// two arrays
for (int i =0; i < (oCount <= eCount?
2*oCount : 2*eCount ); i++ )
{
if( i%2 == 0)
resArr[i] = evenArr[i/2];
else
resArr[i] = oddArr[i/2];
}
// populate rest of elements of max array
// and add zeroes
if (eCount > oCount)
{
for (int i=2*oCount,j=0;i<2*eCount-1; i++)
{
if (i%2 == 0)
{
resArr[i] = evenArr[oCount+j];
j++;
}
else
resArr[i] = 0;
}
}
else if (eCount < oCount)
{
for (int i=2*eCount,j=0;i<2*oCount; i++)
{
if ( i%2 != 0)
{
resArr[i] = oddArr[eCount+j];
j++;
}
else
resArr[i] = 0;
}
}
return resArr;
}
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