JAVA-选择排序从控制台输入数组
2015-10-15 15:46
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原理:a 将数组中的每个元素,与第一个元素比较
如果这个元素小于第一个元素, 就将这个
两个元素交换.
b 每轮使用a的规则, 可以选择出一个最小元素
放到第一个位置.
c 经过n-1轮比较完成排序
简单说: 每轮选择最小的放到前面.
public class selectSort {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("请输入数组以空格分开:");
String str = sc.nextLine();
String[] strs = str.trim().split("\\s{1,}");
int[] ary = new int[strs.length];
for (int i = 0; i < strs.length; i++) {
ary[i]= Integer.parseInt(strs[i]);
}
ary = select(ary);
String s = Arrays.toString(ary);
System.out.println(select(ary));
}
public static int[] select(int[] ary){
for(int i=0;i<ary.length-1;i++){
for(int j=i+1;j<ary.length;j++){
int temp;
if(ary[j]<ary[i]){
temp = ary[i];
ary[i] = ary[j];
ary[j] = temp;
}
}
}
return ary;
}
}
如果这个元素小于第一个元素, 就将这个
两个元素交换.
b 每轮使用a的规则, 可以选择出一个最小元素
放到第一个位置.
c 经过n-1轮比较完成排序
简单说: 每轮选择最小的放到前面.
public class selectSort {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("请输入数组以空格分开:");
String str = sc.nextLine();
String[] strs = str.trim().split("\\s{1,}");
int[] ary = new int[strs.length];
for (int i = 0; i < strs.length; i++) {
ary[i]= Integer.parseInt(strs[i]);
}
ary = select(ary);
String s = Arrays.toString(ary);
System.out.println(select(ary));
}
public static int[] select(int[] ary){
for(int i=0;i<ary.length-1;i++){
for(int j=i+1;j<ary.length;j++){
int temp;
if(ary[j]<ary[i]){
temp = ary[i];
ary[i] = ary[j];
ary[j] = temp;
}
}
}
return ary;
}
}
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