25、JAVA归并排序

归并排序时间复杂度：最好最坏平均都是nlogn

空间复杂度：n

稳定性：稳定的

 

归并排序的用途：

1、排序

　　（速度仅次于快速排序，但较稳定）

2、求逆序对数【这个单独出个笔试题，求逆序对数】

　　具体思路是，在归并的过程中计算每个小区间的逆序对数，进而计算出大区间的逆序对数（也可以用树状数组来求解）

 

JAVA实现归并排序：

 

import java.util.Arrays;public class MergeSort {public static void main(String args[]){int a[] = {4,4,1,6,3};mergeSort(a,0,a.length-1);/*for(int i=0;i<a.length-1;i++)System.out.println(a[i]);*/  System.out.println(Arrays.toString(a));}private static void mergeSort(int a[],int start,int end){//int mid = (start+end)/2;//int[] temp = new int[a.length];int mid=(start&end)+((start^end)>>1);if(start<end){//跳出递归，原来忘记写这一句，导致栈溢出mergeSort(a,start,mid);mergeSort(a,mid+1,end);merge(a,start,mid,mid+1,end);}}private static void merge(int a[],int start1,int end1,int start2,int end2){int i = start1;int j = start2;//int k = 0;//k与此处保持一致，System.arraycopy(temp, 0, a, start1, end2-start1+1);int k = start1;//k与此处保持一致，System.arraycopy(temp, start1, a, start1, end2-start1+1);int temp[] = new int[a.length+1];while(i<=end1 && j<=end2){if(a[i] < a[j]){temp[k++] = a[i++];//temp[k] = a[i];//i++;}else{temp[k++] = a[j++];//temp[k] = a[j];//j++;}//k++;}while(i<=end1){temp[k++] = a[i++];}while(j<=end2){temp[k++] = a[j++];}//k = start1;//for(int element:temp){//把临时数组元素赋值给原数组//a[k++] = element;System.arraycopy(temp, start1, a, start1, end2-start1+1);}}

 

 

删除注释后的完整可运行代码：

import java.util.Arrays;public class MergeSort {public static void main(String args[]){int a[] = {4,4,1,6,3};mergeSort(a,0,a.length-1);/*for(int i=0;i<a.length-1;i++)System.out.println(a[i]);*/  System.out.println(Arrays.toString(a));}private static void mergeSort(int a[],int start,int end){//int mid = (start+end)/2;int mid=(start&end)+((start^end)>>1);if(start<end){//跳出递归，原来忘记写这一句，导致栈溢出mergeSort(a,start,mid);mergeSort(a,mid+1,end);merge(a,start,mid,mid+1,end);}}private static void merge(int a[],int start1,int end1,int start2,int end2){int i = start1;int j = start2;int k = start1;//k与此处保持一致，System.arraycopy(temp, start1, a, start1, end2-start1+1);int temp[] = new int[a.length+1];while(i<=end1 && j<=end2){if(a[i] < a[j]){temp[k++] = a[i++];}else{temp[k++] = a[j++];}}while(i<=end1){temp[k++] = a[i++];}while(j<=end2){temp[k++] = a[j++];}System.arraycopy(temp, start1, a, start1, end2-start1+1);//从temp数组复制到a数组}}