数组中重复的数字

解决问题思路1.

　　

代码实现：

package j2;import java.util.Arrays;/*** Created by admin on 2019/5/8.*/
public class FindDuplicate {public static void duplicate(int[] numbers,int length,int[]duplication){//边界条件的判断if (numbers == null || length==0) {return ;}Arrays.sort(numbers);int j=0;for (int i=0;i<length-1;i++) {if (numbers[i] == numbers[i+1]) {duplication[j] = numbers[i];j++;}}}public static void main(String[] args) {int[] arr={2,3,1,0,2,5,3};int[] duplicate = new int[arr.length];duplicate(arr,arr.length,duplicate);System.out.println(Arrays.toString(duplicate));}}

　　解题思路2.

代码实现：

解题思路3.

代码实现：

package j2;import java.util.Arrays;/*** Created by admin on 2019/5/8.*/
public class FindDuplicate2 {/**** 推荐的做法是：通过交换元素将i保存到numbers[i]中* 在numbers[i]不和i相等的时候，如果number[i]和numbers[numbers[i]]相等就说明是重复的元素* 否则就交换这两个元素，这个过程就是相当于排序* @param numbers* @param length* @param duplication*/public static void duplicate(int[] numbers,int length,int[]duplication){if (numbers == null || length<= 0) {return ;}for (int i=0;i<length;i++) {if (numbers[i]<0 || numbers[i]>length-1) {return;}}//如果i下标和i所指定的元素不同，那么我们就要进行交换，否则再比较要找的元素是否和i所指定的元素相同，如果是就是重复的元素int k=0;for (int i=0;i<length;i++) {while (numbers[i] != i){//现在numbers[i] != i,设numbers[i]=j,所以如果下面的if成立，就是numbers[i] = numbers[j],说明重复if (numbers[i] == numbers[numbers[i]]) {duplication[k] = numbers[i];k++;break;}//交换元素swap(numbers,i,numbers[i]);}}}//交换元素private static void swap(int[] numbers, int p, int q) {int temp = numbers[p];numbers[p] = numbers[q];numbers[q] = temp;}public static void main(String[] args) {int[] arr={2,3,1,0,2,5,3};int[] duplicate = new int[arr.length];duplicate(arr,arr.length,duplicate);System.out.println(Arrays.toString(duplicate));}}