1. 選擇排序
選擇排序的基本思想是周遊數組的過程中,以i代表目前需要排序的序号,則需要在剩餘的[i..n-1]中找出其中的最小值,然後将找到的最小值與i指向的值進行交換。因為每一次确定元素的過程中都會有一個選擇很大值的子流程,是以稱之為選擇排序。
比如[38, 17, 16, 16, 7, 31, 39, 32, 2, 11]
i = 0: [2 , 17, 16, 16, 7, 31, 39, 32, 38 , 11] (0th [38]<->8th [2])
i = 1: [2, 7 , 16, 16, 17 , 31, 39, 32, 38, 11] (1st [38]<->4th [17])
i = 2: [2, 7, 11 , 16, 17, 31, 39, 32, 38, 16 ] (2nd [11]<->9th [16])
i = 3: [2, 7, 11, 16, 17, 31, 39, 32, 38, 16] ( 無需交換 )
i = 4: [2, 7, 11, 16, 16 , 31, 39, 32, 38, 17 ] (4th [17]<->9th [16])
i = 5: [2, 7, 11, 16, 16, 17 , 39, 32, 38, 31 ] (5th [31]<->9th [17])
i = 6: [2, 7, 11, 16, 16, 17, 31 , 32, 38, 39 ] (6th [39]<->9th [31])
i = 7: [2, 7, 11, 16, 16, 17, 31, 32, 38, 39] ( 無需交換 )
i = 8: [2, 7, 11, 16, 16, 17, 31, 32, 38, 39] ( 無需交換 )
i = 9: [2, 7, 11, 16, 16, 17, 31, 32, 38, 39] ( 無需交換 )
選擇排序随着排序的進行(i逐漸增大),比較的次數會越來越少,但是不論數組初始是否有序,選擇排序都會從i至數組末尾進行一次選擇比較,是以給定長度的數組,選擇排序的比較次數是固定的:1+2+3+…+n=n*(n+1)/2,而交換的次數則跟初始數組的順序有關,如果初始數組順序為随機,則在最壞情況下數組元素将會交換N次,最好的情況是0次。選擇排序的時間複雜度和空間複雜度分别為O(n2 ) 和 O(1)。
package algorithms;
publicabstractclass Sorter<E extends Comparable<E>> {
publicabstractvoid sort(E[] array,int from ,int len);
publicfinalvoid sort(E[] array)
{
sort(array,0,array.length);
}
protectedfinalvoid swap(E[] array,int from ,int to)
E tmp=array[from];
array[from]=array[to];
array[to]=tmp;
publicvoid sort(String helloString, int from, int len) {
// TODO Auto-generated method stub
}
}
/**
* @author yovn
*
*/
publicclass SelectSorter<E extends Comparable<E>> extends Sorter<E> {
/* (non-Javadoc)
* @see algorithms.Sorter#sort(E[], int, int)
*/
@Override
publicvoid sort(E[] array, int from, int len) {
for(int i=0;i<len;i++)
{
int smallest=i;
int j=i+from;
for(;j<from+len;j++)
{
if(array[j].compareTo(array[smallest])<0)
{
smallest=j;
}
}
swap(array,i,smallest);
}
publicstaticvoid main(String[] args){
String[] myStringArray = new String[3];
String[] myStringArray1 = {"a","c","b"};
String[] myStringArray2 = new String[]{"a","b","c"};
SelectSorter<String> s1 = new SelectSorter<String>();
for(int i=0;i<3;i++){
System.out.println(myStringArray1[i]);
}
s1.sort(myStringArray1, 0, 3);
Output:
a
c
b