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Merge Sort

时间:2017-03-12 11:24:54      阅读:209      评论:0      收藏:0      [点我收藏+]

标签:lag   rdo   als   orm   new   better   nbsp   efi   with   

character:the sort time is proportional to any input array with length of N.

Introduction to merge:if two arrays are already sorted,after merge operation,the combined array is also sorted.

the basic realization of merge function:

public static void merge(Comparable[] a,int lo,int hi){
if(hi<=lo) return;
boolean flag=true;

int mid=lo+(hi-lo)/2;
int head=lo;
int tail=mid+1;
int index=lo;

for(int i=lo;i<=hi;++i){
aux[i]=a[i];
}
while(flag){
boolean inHead= (head<=mid);
boolean inTail= (tail<=hi);

if((head<=mid) && (tail<=hi)){
if(less(aux[head],aux[tail])){
a[index++]=aux[head++];
}else{
a[index++]=aux[tail++];
}
}

if((inHead) && (!inTail)){
while(head<=mid){
a[index++]=aux[head++];
}
}

if((!inHead) && (inTail)){
while(tail<=hi){
a[index++]=aux[tail++];
}
}

if((!inHead) && (!inTail)){
flag=false;
}
}
}

the update one:

public static void updateMerge(Comparable[] a,int lo,int hi){
int mid=lo+(hi-lo)/2;
int index=lo;
int head=lo;
int tail=mid+1;

for(int i=lo;i<=hi;++i){
aux[i]=a[i];
}

while(index<=hi){
if(head>mid){a[index++]=aux[tail++];}
else if(tail>hi){a[index++]=aux[head++];}
else if(less(aux[head],aux[tail])){a[index++]=aux[head++];}
else{a[index++]=aux[tail++];}
}
}

you may notice the aux array,it‘s defined as an inner variable of MergeSort class. 

then it‘s time to sort a given array,we normally have two methods:the up2Down one and the down2Up one.

the up2Down realization:

public static void upDownMergeSort(Comparable[] a,int lo,int hi){
if(hi<=lo) return;
int mid=lo+(hi-lo)/2;
upDownMergeSort(a,lo,mid);
upDownMergeSort(a,mid+1,hi);
updateMerge(a,lo,hi);
}

the down2Up realization:

also this way has two methods:the complex method and a clear method

at first ,I‘ll introduce the complex method.it‘s thinking is simpler ,so the numbers of code is greater.

when the length of input array is the power of 2,the calculation is easy.the code is in the below:

public static void innerDownUpMergeSort(Comparable[] a,int lo,int hi){
if(hi<=lo) return;
int len=hi-lo+1;
int step=2;
aux=new Comparable[len];
int stages=0;
while(step<=len){
stages=Math.floorDiv(len,step);
for(int i=0;i<stages;++i){
merge(a,lo+i*step,lo+(i+1)*step-1);
}
step*=2;
}
}

it works well when the input array length is 2‘s power,so we think that may be we can do something to change the common array into the special one with the length of 2^N. follow this way we can code as below:

public static void downUpMergeSort(Comparable[] a,int lo,int hi){
if(hi<=lo) return;
int len=hi-lo+1;

int N=len;
if(0 != is2sPow((double)len)) {
double lower = Math.log((double) len) / Math.log((double) 2);
lower = Math.floor(lower)+1;
N=(int)Math.pow(2,lower);
}

Comparable[] innerAux=new Comparable[N];
for(int i=lo;i<=hi;++i){
innerAux[i]=a[i];
}
for(int i=hi+1;i<N;++i){
innerAux[i]=MAXINT;
}

innerDownUpMergeSort(innerAux,0,N-1);

for(int i=lo;i<=hi;++i){
a[i]=innerAux[i];
}
}

then there is a function called is2sPow(),the explanation is showed in http://www.cnblogs.com/ssMellon/p/6423101.html

it works ,but it‘s not clear.we can do it better.but unfortunately I found a function in a book called Algorithm decribed by Java.

the following is the code:

public static void updateDownUpMergeSort(Comparable[] a){
int N=a.length;
aux=new Comparable[N];
for(int sz=1;sz<N;sz*=2){
for(int lo=0;lo<N-sz;lo+=2*sz){
merge(a,lo,Math.min(lo+2*sz-1,N-1));
}
}
}

So clear.

The end.

Merge Sort

标签:lag   rdo   als   orm   new   better   nbsp   efi   with   

原文地址:http://www.cnblogs.com/ssMellon/p/6536966.html

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