标签:default style tin list rem 序号 实现 quicksort false
1:简单选择排序
2:堆排序
3:冒泡排序
4:归并排序
5:基数排序
6:快速排序
package cn.sort; import java.util.Scanner; //各种控排序算法 public class Test { public static void main(String[] args) { int array[] = {22, 11, 9, 8,90, 7, 42}; System.out.println("没有排序之前:"); Sortall.print(array); System.out.println(); int num =1; System.out.println("请输入需要排序的方法序号:"); System.out.println("1:简单选择排序"); System.out.println("2:堆排序"); System.out.println("3:冒泡排序"); System.out.println("4:归并排序"); System.out.println("5:基数排序"); System.out.println("6:快速排序"); System.out.println("0:退出程序"); boolean loop = true; while(loop){ Scanner sc = new Scanner(System.in); num = sc.nextInt(); switch(num){ case 0: loop=false; System.out.println("退出程序"); break; case 1://简单选择排序 Sortall.SelectSort(array); System.out.println(); break; case 2: //堆排序 Sortall.HeapSort(array); System.out.println(); break; case 3: //冒泡排序 Sortall.BubbleSort(array); System.out.println(); break; case 4: //归并排序 Sortall.MergeSort(array); System.out.println(); break; case 5: //基数排序 Sortall.BaseSort(array); System.out.println(); break; case 6: //快速排序 Sortall.QuickSort(array); System.out.println(); break; default: System.out.println("无效输入\n"); break; } } } }
方法实现:
package cn.sort; import java.util.ArrayList; import java.util.List; public class Sortall{ //重写print方法输出数组 static void print(int array[]) { for (int i = 0; i < array.length; i++) { System.out.print(array[i]+" "); } } //交换函数 static void Swap(int a [],int i,int j){ int temp = a[i]; a[i] = a[j]; a[j] = temp; } //1.选择排序 public static void SelectSort(int []array){ System.out.println("----------------1.使用选择排序之后: ----------------------"); for(int i=0;i<array.length;i++){ int lowindex=i; for(int j=i+1;j<array.length;j++){ if(array[j]<array[lowindex]){ lowindex=j; } } int temp=array[i]; array[i]=array[lowindex]; array[lowindex]=temp; } print(array); } //-------------------------另一种选择排序的写方法-------------------------------------------------------- public static void xuanZe1(int []array){ for(int i=0;i<array.length;i++){ for(int j=i+1;j<array.length;j++){ if(array[j]<array[i]){ Swap(array, i, j); } } } for(int i=0;i<array.length;i++){ System.out.print(array[i]+" "); } } //2.二分法查找数据 public static int BinarySearch(int[] array, int num) { int begin = 0; int end = array.length-1; while (begin <= end) { int mid = (int)Math.floor((begin+end)/2); for (int i = 0; i < array.length; i++) { if (array[mid] == num) { System.out.println("在数组中找到"+num+"的位置为:"+mid); return mid; }else if (array[mid] < num) { begin = mid +1; }else{ end = mid -1; } } } return -1; } //3.冒泡排序 public static void BubbleSort(int[] array) { System.out.println("----------------3.使用冒泡排序之后: -----------------------"); for (int i = 0; i < array.length - 1; i++) { for (int j = array.length - 1; j >i ; j--) { if (array[j]<array[j-1]) { Swap(array, j, j-1); } } } } //堆排序 public static void HeapSort(int[] array) { System.out.println("----------------2.使用堆排序之后: -----------------------"); //循环建堆 for (int i = 0; i < array.length; i++) { buidMaxHeap(array, array.length - 1 - i); //交换堆顶和最后一个元素 Swap(array, 0, array.length-i-1); } System.out.println("输出排序:"); print(array); } //建堆 private static void buidMaxHeap(int[] data,int lastIndex) { //从lastIndex处最后一个节点的父节点开始 for (int i = (lastIndex - 1)/2; i >= 0; i--) { //k保存正在判断的节点 int k = i; //如果当前k节点的子节点存在 while (k*2+1 <= lastIndex) { //k节点的左子节点的索引 int biggerIndex = 2*k + 1; //如果biggerIndex小于lastIndex,即biggerIndex代表的k节点的右子节点存在 if (biggerIndex < lastIndex) { //如果右子节点的值较大 if (data[biggerIndex] < data[biggerIndex + 1]) { //biggerIndex总是记录较大子节点的索引 biggerIndex++; } } //如果k节点的值小于其较大的子节点的值 if (data[k]<data[biggerIndex]) { //交换他们 Swap(data, k, biggerIndex); }else{ break; } } } print(data); } //快速排序 public static void QuickSort(int[] array) { System.out.println("----------------6.使用快速排序之后: -----------------------"); if (array.length>0) { quick(array, 0, array.length-1); } print(array); } static void quick(int a[], int low, int high){ if (low < high) { int middle = getMiddle(a, low, high); quick(a, 0, middle-1); quick(a, middle+1, high); } } static int getMiddle(int a[],int low,int high){ int temp = a[low]; while(low < high){ while(low < high && a[low]<= temp){ low ++; } a[high] = a[low]; } a[low] = temp; return low; } //归并排序 public static void MergeSort(int[] array) { System.out.println("----------------4.使用归并排序之后: -----------------------"); if (array.length>0) { mergeSort(array, 0, array.length - 1); } print(array); } private static void mergeSort(int[] a, int left, int right) { if(left<right){ int middle = (left+right)/2; //对左边进行递归 mergeSort(a, left, middle); //对右边进行递归 mergeSort(a, middle+1, right); //合并 merge(a,left,middle,right); } } private static void merge(int[] a, int left, int middle, int right) { int[] tmpArr = new int[a.length]; int mid = middle+1; //右边的起始位置 int tmp = left; int third = left; while(left<=middle && mid<=right){ //从两个数组中选取较小的数放入中间数组 if(a[left]<=a[mid]){ tmpArr[third++] = a[left++]; }else{ tmpArr[third++] = a[mid++]; } } //将剩余的部分放入中间数组 while(left<=middle){ tmpArr[third++] = a[left++]; } while(mid<=right){ tmpArr[third++] = a[mid++]; } //将中间数组复制回原数组 while(tmp<=right){ a[tmp] = tmpArr[tmp++]; } } //基数排序 public static void BaseSort(int[] array) { System.out.println("----------------5.使用基数排序之后: -----------------------"); //找到最大数,确定要排序几趟 int max = 0; for (int i = 0; i < array.length; i++) { if(max<array[i]){ max = array[i]; } } //判断位数 int times = 0; while(max>0){ max = max/10; times++; } //建立十个队列 List<ArrayList> queue = new ArrayList<ArrayList>(); for (int i = 0; i < 10; i++) { ArrayList queue1 = new ArrayList(); queue.add(queue1); } //进行times次分配和收集 for (int i = 0; i < times; i++) { //分配 for (int j = 0; j < array.length; j++) { int x = array[j]%(int)Math.pow(10, i+1)/(int)Math.pow(10, i); ArrayList queue2 = queue.get(x); queue2.add(array[j]); queue.set(x,queue2); } //收集 int count = 0; for (int j = 0; j < 10; j++) { while(queue.get(j).size()>0){ ArrayList<Integer> queue3 = queue.get(j); array[count] = queue3.get(0); queue3.remove(0); count++; } } } print(array); } }
标签:default style tin list rem 序号 实现 quicksort false
原文地址:http://www.cnblogs.com/mengjie1001/p/6846262.html