Java常用排序算法/程序员必须掌握的8大排序算法
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分類:
1)插入排序(直接插入排序、希爾排序)
2)交換排序(冒泡排序、快速排序)
3)選擇排序(直接選擇排序、堆排序)
4)歸并排序
5)分配排序(基數(shù)排序)
所需輔助空間最多:歸并排序
所需輔助空間最少:堆排序
平均速度最快:快速排序
不穩(wěn)定:快速排序,希爾排序,堆排序。
先來看看8種排序之間的關(guān)系:
??
?1.直接插入排序
(1)基本思想:在要排序的一組數(shù)中,假設(shè)前面(n-1)[n>=2] 個(gè)數(shù)已經(jīng)是排
好順序的,現(xiàn)在要把第n個(gè)數(shù)插到前面的有序數(shù)中,使得這n個(gè)數(shù)
也是排好順序的。如此反復(fù)循環(huán),直到全部排好順序。
(2)實(shí)例
(3)用java實(shí)現(xiàn)
?
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 | package com.njue;? ??? publicclass insertSort {? ??? public insertSort(){? ????inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,56,17,18,23,34,15,35,25,53,51};? ????int temp=0;? ????for(int i=1;i<a.length;i++){? ???????int j=i-1;? ???????temp=a[i];? ???????for(;j>=0&&temp<a[j];j--){? ???????????a[j+1]=a[j];? //將大于temp的值整體后移一個(gè)單位? ???????}? ???????a[j+1]=temp;? ????}? ??? ????for(int i=0;i<a.length;i++){? ???????System.out.println(a[i]);? ????}? } |
2.?? 希爾排序(最小增量排序)
(1)基本思想:算法先將要排序的一組數(shù)按某個(gè)增量d(n/2,n為要排序數(shù)的個(gè)數(shù))分成若干組,每組中記錄的下標(biāo)相差d.對每組中全部元素進(jìn)行直接插入排序,然后再用一個(gè)較小的增量(d/2)對它進(jìn)行分組,在每組中再進(jìn)行直接插入排序。當(dāng)增量減到1時(shí),進(jìn)行直接插入排序后,排序完成。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
?
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | publicclass shellSort {? ??? publicshellSort(){? ??? ????int a[]={1,54,6,3,78,34,12,45,56,100};? ????double d1=a.length;? ????int temp=0;? ??? ????while(true){? ???????d1= Math.ceil(d1/2);? ???????int d=(int) d1;? ???????for(int x=0;x<d;x++){? ??? ???????????for(int i=x+d;i<a.length;i+=d){? ??????????????int j=i-d;? ??????????????temp=a[i];? ??????????????for(;j>=0&&temp<a[j];j-=d){? ???????????????????a[j+d]=a[j];? ??????????????}? ??????????????a[j+d]=temp;? ???????????}? ???????}? ??? ???????if(d==1){? ???????????break;? ???????}? ??? ????for(int i=0;i<a.length;i++){? ???????System.out.println(a[i]);? ????}? } |
3.簡單選擇排序
(1)基本思想:在要排序的一組數(shù)中,選出最小的一個(gè)數(shù)與第一個(gè)位置的數(shù)交換;
然后在剩下的數(shù)當(dāng)中再找最小的與第二個(gè)位置的數(shù)交換,如此循環(huán)到倒數(shù)第二個(gè)數(shù)和最后一個(gè)數(shù)比較為止。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | public class selectSort {? ??? ????public selectSort(){? ???????int a[]={1,54,6,3,78,34,12,45};? ???????int position=0;? ???????for(int i=0;i<a.length;i++){?????? ???????????int j=i+1;? ???????????position=i;? ???????????int temp=a[i];? ???????????for(;j<a.length;j++){? ??????????????if(a[j]<temp){? ?????????????????temp=a[j];? ?????????????????position=j;? ??????????????}? ???????????}? ???????????a[position]=a[i];? ???????????a[i]=temp;? ???????}? ??? ???????for(int i=0;i<a.length;i++)? ???????????System.out.println(a[i]);? ????}? } |
4.堆排序
(1)基本思想:堆排序是一種樹形選擇排序,是對直接選擇排序的有效改進(jìn)。
堆的定義如下:具有n個(gè)元素的序列(h1,h2,…,hn),當(dāng)且僅當(dāng)滿足(hi>=h2i,hi>=2i+1)或(hi<=h2i,hi<=2i+1)(i=1,2,…,n/2)時(shí)稱之為堆。在這里只討論滿足前者條件的堆。由堆的定義可以看出,堆頂元素(即第一個(gè)元素)必為最大項(xiàng)(大頂堆)。完全二叉樹可以很直觀地表示堆的結(jié)構(gòu)。堆頂為根,其它為左子樹、右子樹。初始時(shí)把要排序的數(shù)的序列看作是一棵順序存儲的二叉樹,調(diào)整它們的存儲序,使之成為一個(gè)堆,這時(shí)堆的根節(jié)點(diǎn)的數(shù)最大。然后將根節(jié)點(diǎn)與堆的最后一個(gè)節(jié)點(diǎn)交換。然后對前面(n-1)個(gè)數(shù)重新調(diào)整使之成為堆。依此類推,直到只有兩個(gè)節(jié)點(diǎn)的堆,并對它們作交換,最后得到有n個(gè)節(jié)點(diǎn)的有序序列。從算法描述來看,堆排序需要兩個(gè)過程,一是建立堆,二是堆頂與堆的最后一個(gè)元素交換位置。所以堆排序有兩個(gè)函數(shù)組成。一是建堆的滲透函數(shù),二是反復(fù)調(diào)用滲透函數(shù)實(shí)現(xiàn)排序的函數(shù)。
(2)實(shí)例:
初始序列:46,79,56,38,40,84
建堆:
交換,從堆中踢出最大數(shù)
剩余結(jié)點(diǎn)再建堆,再交換踢出最大數(shù)
依次類推:最后堆中剩余的最后兩個(gè)結(jié)點(diǎn)交換,踢出一個(gè),排序完成。
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 | import java.util.Arrays;? ??? publicclass HeapSort {? ????inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,56,17,18,23,34,15,35,25,53,51};? ????public? HeapSort(){? ???????heapSort(a);? ????}? ??? ????public? void heapSort(int[] a){? ????????System.out.println("開始排序");? ????????int arrayLength=a.length;? ????????//循環(huán)建堆? ????????for(int i=0;i<arrayLength-1;i++){? ????????????//建堆? ????????????buildMaxHeap(a,arrayLength-1-i);? ????????????//交換堆頂和最后一個(gè)元素? ????????????swap(a,0,arrayLength-1-i);? ????????????System.out.println(Arrays.toString(a));? ????????}? ????}? ??? ???? ??? ????private? void swap(int[] data, int i, int j) {? ????????// TODO Auto-generated method stub? ????????int tmp=data[i];? ????????data[i]=data[j];? ????????data[j]=tmp;? ????}? ??? ????//對data數(shù)組從0到lastIndex建大頂堆? ????privatevoid buildMaxHeap(int[] data, int lastIndex) {? ????????// TODO Auto-generated method stub? ????????//從lastIndex處節(jié)點(diǎn)(最后一個(gè)節(jié)點(diǎn))的父節(jié)點(diǎn)開始? ??? ????????for(int i=(lastIndex-1)/2;i>=0;i--){? ????????????//k保存正在判斷的節(jié)點(diǎn)? ????????????int k=i;? ????????????//如果當(dāng)前k節(jié)點(diǎn)的子節(jié)點(diǎn)存在? ????????????while(k*2+1<=lastIndex){? ????????????????//k節(jié)點(diǎn)的左子節(jié)點(diǎn)的索引? ????????????????int biggerIndex=2*k+1;? ????????????????//如果biggerIndex小于lastIndex,即biggerIndex+1代表的k節(jié)點(diǎn)的右子節(jié)點(diǎn)存在? ????????????????if(biggerIndex<lastIndex){? ????????????????????//若果右子節(jié)點(diǎn)的值較大? ????????????????????if(data[biggerIndex]<data[biggerIndex+1]){? ????????????????????????//biggerIndex總是記錄較大子節(jié)點(diǎn)的索引? ????????????????????????biggerIndex++;? ????????????????????}? ????????????????}? ??? ????????????????//如果k節(jié)點(diǎn)的值小于其較大的子節(jié)點(diǎn)的值? ???????????????if(data[k]<data[biggerIndex]){? ????????????????????//交換他們? ????????????????????swap(data,k,biggerIndex);? ????????????????????//將biggerIndex賦予k,開始while循環(huán)的下一次循環(huán),重新保證k節(jié)點(diǎn)的值大于其左右子節(jié)點(diǎn)的值? ????????????????????k=biggerIndex;? ????????????????}else{? ????????????????????break;? ????????????????}? ????????????}? ????????}? ????}? } |
5.冒泡排序
(1)基本思想:在要排序的一組數(shù)中,對當(dāng)前還未排好序的范圍內(nèi)的全部數(shù),自上而下對相鄰的兩個(gè)數(shù)依次進(jìn)行比較和調(diào)整,讓較大的數(shù)往下沉,較小的往上冒。即:每當(dāng)兩相鄰的數(shù)比較后發(fā)現(xiàn)它們的排序與排序要求相反時(shí),就將它們互換。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | public class bubbleSort {? ??? publicbubbleSort(){? ?????inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,56,17,18,23,34,15,35,25,53,51};? ????int temp=0;? ????for(int i=0;i<a.length-1;i++){? ???????for(int j=0;j<a.length-1-i;j++){? ?????????if(a[j]>a[j+1]){? ???????????temp=a[j];? ???????????a[j]=a[j+1];? ???????????a[j+1]=temp;? ?????????}? ???????}? ????}? ??? ????for(int i=0;i<a.length;i++){? ???????System.out.println(a[i]);??? ???}? } |
6.快速排序
(1)基本思想:選擇一個(gè)基準(zhǔn)元素,通常選擇第一個(gè)元素或者最后一個(gè)元素,通過一趟掃描,將待排序列分成兩部分,一部分比基準(zhǔn)元素小,一部分大于等于基準(zhǔn)元素,此時(shí)基準(zhǔn)元素在其排好序后的正確位置,然后再用同樣的方法遞歸地排序劃分的兩部分。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 | publicclass quickSort {? ??? ??inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,56,17,18,23,34,15,35,25,53,51};? publicquickSort(){? ????quick(a);? ????for(int i=0;i<a.length;i++){? ???????System.out.println(a[i]);? ????}? }? publicint getMiddle(int[] list, int low, int high) {??? ????????????int tmp =list[low];??? //數(shù)組的第一個(gè)作為中軸??? ????????????while (low < high){??? ????????????????while (low < high&& list[high] >= tmp) {??? ???????????????????high--;??? ????????????????}??? ??? ????????????????list[low] =list[high];?? //比中軸小的記錄移到低端??? ????????????????while (low < high&& list[low] <= tmp) {??? ????????????????????low++;??? ????????????????}??? ??? ????????????????list[high] =list[low];?? //比中軸大的記錄移到高端??? ????????????}??? ???????????list[low] = tmp;????????????? //中軸記錄到尾??? ????????????return low;?????????????????? //返回中軸的位置??? }?? ??? publicvoid _quickSort(int[] list, int low, int high) {??? ????????????if (low < high){??? ???????????????int middle =getMiddle(list, low, high);? //將list數(shù)組進(jìn)行一分為二??? ???????????????_quickSort(list, low, middle - 1);?????? //對低字表進(jìn)行遞歸排序??? ???????????????_quickSort(list,middle + 1, high);?????? //對高字表進(jìn)行遞歸排序??? ????????????}??? }? ??? publicvoid quick(int[] a2) {??? ????????????if (a2.length > 0) {??? //查看數(shù)組是否為空??? ????????????????_quickSort(a2,0, a2.length - 1);??? ????????????}??? }? } |
7、歸并排序
(1)基本排序:歸并(Merge)排序法是將兩個(gè)(或兩個(gè)以上)有序表合并成一個(gè)新的有序表,即把待排序序列分為若干個(gè)子序列,每個(gè)子序列是有序的。然后再把有序子序列合并為整體有序序列。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 | import java.util.Arrays;? ??? publicclass mergingSort {? ??? inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,56,17,18,23,34,15,35,25,53,51};? ??? publicmergingSort(){? ????sort(a,0,a.length-1);? ????for(int i=0;i<a.length;i++)? ???????System.out.println(a[i]);? }? ??? publicvoid sort(int[] data, int left, int right) {? ????// TODO Auto-generatedmethod stub? ????if(left<right){? ????????//找出中間索引? ????????int center=(left+right)/2;? ????????//對左邊數(shù)組進(jìn)行遞歸? ????????sort(data,left,center);? ????????//對右邊數(shù)組進(jìn)行遞歸? ????????sort(data,center+1,right);? ????????//合并? ????????merge(data,left,center,right);???????? ????}? ??? }? ??? publicvoid merge(int[] data, int left, int center, int right) {? ????// TODO Auto-generatedmethod stub? ????int [] tmpArr=newint[data.length];? ????int mid=center+1;? ????//third記錄中間數(shù)組的索引? ????int third=left;? ????int tmp=left;? ????while(left<=center&&mid<=right){? ????????//從兩個(gè)數(shù)組中取出最小的放入中間數(shù)組? ????????if(data[left]<=data[mid]){? ????????????tmpArr[third++]=data[left++];? ????????}else{? ????????????tmpArr[third++]=data[mid++];? ????????}? ??? ????}? ??? ????//剩余部分依次放入中間數(shù)組? ????while(mid<=right){? ????????tmpArr[third++]=data[mid++];? ????}? ??? ????while(left<=center){? ????????tmpArr[third++]=data[left++];? ????}? ??? ????//將中間數(shù)組中的內(nèi)容復(fù)制回原數(shù)組? ????while(tmp<=right){? ????????data[tmp]=tmpArr[tmp++];? ????}? ????System.out.println(Arrays.toString(data));? }? } |
8、基數(shù)排序
(1)基本思想:將所有待比較數(shù)值(正整數(shù))統(tǒng)一為同樣的數(shù)位長度,數(shù)位較短的數(shù)前面補(bǔ)零。然后,從最低位開始,依次進(jìn)行一次排序。這樣從最低位排序一直到最高位排序完成以后,數(shù)列就變成一個(gè)有序序列。
(2)實(shí)例:
(3)用java實(shí)現(xiàn)
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 | import java.util.ArrayList;? import java.util.List;? ??? public class radixSort {? ????inta[]={49,38,65,97,76,13,27,49,78,34,12,64,5,4,62,99,98,54,101,56,17,18,23,34,15,35,25,53,51};? ????public radixSort(){? ???????sort(a);? ???????for(inti=0;i<a.length;i++){? ??????????????System.out.println(a[i]);? ???????}? ????}???????? ????public? void sort(int[] array){??? ???????//首先確定排序的趟數(shù);??? ???????int max=array[0];??? ???????for(inti=1;i<array.length;i++){??? ????????????if(array[i]>max){??? ??????????????max=array[i];??? ????????????}??? ???????}??? ???????int time=0;??? ???????//判斷位數(shù);??? ???????while(max>0){??? ??????????max/=10;??? ???????????time++;??? ???????}??? ??? ????????//建立10個(gè)隊(duì)列;??? ???????List<ArrayList> queue=newArrayList<ArrayList>();??? ???????for(int i=0;i<10;i++){??? ??????????????ArrayList<Integer>queue1=new ArrayList<Integer>();? ???????????queue.add(queue1);??? ???????}??? ??? ???????//進(jìn)行time次分配和收集;??? ???????for(int i=0;i<time;i++){??? ???????????//分配數(shù)組元素;??? ??????????for(intj=0;j<array.length;j++){??? ???????????????//得到數(shù)字的第time+1位數(shù);? ?????????????????int x=array[j]%(int)Math.pow(10,i+1)/(int)Math.pow(10, i);? ?????????????????ArrayList<Integer>queue2=queue.get(x);? ?????????????????queue2.add(array[j]);? ?????????????????queue.set(x, queue2);? ??????????}?? ??????????int count=0;//元素計(jì)數(shù)器;??? ??????????//收集隊(duì)列元素;??? ??????????for(int k=0;k<10;k++){? ???????????????while(queue.get(k).size()>0){? ???????????????????ArrayList<Integer>queue3=queue.get(k);? ???????????????????array[count]=queue3.get(0);??? ???????????????????queue3.remove(0);? ???????????????????count++;? ???????????????}?? ??????????}??? ???????}?????????????? ????}? } |
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