1. 选择排序
选择排序的基本思想是遍历数组的过程中，以i代表当前需要排序的序号，则需要在剩余的[i..n-1]中找出其中的最小值，然后将找到的最小值与i指向的值进行交换。因为每一次确定元素的过程中都会有一个选择很大值的子流程，所以称之为选择排序。
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] ( 无需交换 )
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] ( 无需交换 )
i = 8: [2, 7, 11, 16, 16, 17, 31, 32, 38, 39] ( 无需交换 )
i = 9: [2, 7, 11, 16, 16, 17, 31, 32, 38, 39] ( 无需交换 )
选择排序随着排序的进行(i逐渐增大)，比较的次数会越来越少，但是不论数组初始是否有序，选择排序都会从i至数组末尾进行一次选择比较，所以给定长度的数组，选择排序的比较次数是固定的:1+2+3+…+n=n*(n+1)/2,而交换的次数则跟初始数组的顺序有关，如果初始数组顺序为随机，则在最坏情况下数组元素将会交换N次，最好的情况是0次。选择排序的时间复杂度和空间复杂度分别为O(n2 ) 和 O(1)。
package algorithms;
publicabstractclass Sorter<Eextends 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
}
}
package algorithms;
/**
* @authoryovn
*
*/
publicclass SelectSorter<Eextends 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);
for(int i=0;i<3;i++){
System.out.println(myStringArray1[i]);
}
}
}
Output:
a
c
b
a
b
c