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MaxMinSelectActivity.java
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MaxMinSelectActivity.java
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package com.wangpos.datastructure.sort;
import com.wangpos.datastructure.core.BaseActivity;
import com.wangpos.datastructure.core.CodeBean;
import java.util.Arrays;
/**
* Created by qiyue on 2017/11/26.
*/
public class MaxMinSelectActivity extends BaseActivity {
int arr[] = {23, 12, 13, 44, 65, 26, 17, 38, 59};
@Override
protected void initData() {
addItem(new CodeBean("同时最大值最小值快速查找法",select_max));
}
@Override
protected String getTextData() {
return Arrays.toString(arr);
}
@Override
protected int getImageData() {
return 0;
}
@Override
protected String getResultData() {
return SelectMax(arr).toString();
}
@Override
protected String getTimeData() {
return "O(n)";
}
@Override
protected String getSpaceTimeData() {
return null;
}
@Override
protected String getWendingXingData() {
return "稳定";
}
@Override
protected String getSummaryData() {
return "复杂度相同,但是比较次数,缺少了很多,这样如果比较比较耗时,并且数据量很大,就会更加有效率";
}
class MaxMin {
public int max;
public int min;
@Override
public String toString() {
return "max = "+max +"," +"min ="+min;
}
}
private static final String select_max = "/**\n" +
" * 如果是奇数 3(n/2)\n" +
" * 如果是偶数 3(n-2)/2\n" +
" *\n" +
" * 他们时间复杂度同样是 n n/2,当然按照无穷大,时间复杂度都是n,但比较次数少的更快些\n" +
" * @param a\n" +
" * @return\n" +
" */\n" +
" public MaxMin SelectMax(int a[]){\n" +
" MaxMin num = new MaxMin();\n" +
"\n" +
" int i = 0;\n" +
" int len = a.length;\n" +
" if (len%2!=0)\n" +
" {\n" +
" num.max = num.min = a[0];\n" +
" i = 1;\n" +
" }\n" +
" else\n" +
" {\n" +
" //如果是偶数,则先比较前两个元素,然后成对处理后面元素。\n" +
" if (a[0]>a[1])\n" +
" {\n" +
" num.max = a[0];\n" +
" num.min = a[1];\n" +
" }\n" +
" else\n" +
" {\n" +
" num.max = a[1];\n" +
" num.min = a[0];\n" +
" }\n" +
" i = 2;\n" +
" }\n" +
"\n" +
" for(;i<len;i+=2){\n" +
"\n" +
" int max_temp;\n" +
" int min_temp;\n" +
" //首先将两个元素进行比较得出较大较小值\n" +
" if (a[i] > a[i + 1])\n" +
" {\n" +
" max_temp = a[i];\n" +
" min_temp = a[i + 1];\n" +
" }else{\n" +
" max_temp = a[i+1];\n" +
" min_temp = a[i];\n" +
" }\n" +
"\n" +
" //然后将较大较小值与最大最小值进行比较\n" +
" num.max = num.max > max_temp ? num.max : max_temp;\n" +
" num.min = num.min < min_temp ? num.min : min_temp;\n" +
" }\n" +
"\n" +
" return num;\n" +
"\n" +
"\n" +
" }\n" +
"\n" +
"\n" +
" //对比\n" +
" public MaxMin oldSelect(int a[]){\n" +
" MaxMin num = new MaxMin();\n" +
"\n" +
" /**\n" +
" * 2n次比较\n" +
" */\n" +
" for(int i = 0;i<a.length;i++){\n" +
" if (a[i]>num.max){\n" +
" num.max = a[i];\n" +
" }\n" +
" if (a[i]<num.min){\n" +
" num.min = a[i];\n" +
" }\n" +
" }\n" +
" return num;\n" +
" }";
/**
* 如果是奇数 3(n/2)
* 如果是偶数 3(n-2)/2
*
* 他们时间复杂度同样是 n n/2,当然按照无穷大,时间复杂度都是n,但比较次数少的更快些
* @param a
* @return
*/
public MaxMin SelectMax(int a[]){
MaxMin num = new MaxMin();
int i = 0;
int len = a.length;
if (len%2!=0)
{
num.max = num.min = a[0];
i = 1;
}
else
{
//如果是偶数,则先比较前两个元素,然后成对处理后面元素。
if (a[0]>a[1])
{
num.max = a[0];
num.min = a[1];
}
else
{
num.max = a[1];
num.min = a[0];
}
i = 2;
}
for(;i<len;i+=2){
int max_temp;
int min_temp;
//首先将两个元素进行比较得出较大较小值
if (a[i] > a[i + 1])
{
max_temp = a[i];
min_temp = a[i + 1];
}else{
max_temp = a[i+1];
min_temp = a[i];
}
//然后将较大较小值与最大最小值进行比较
num.max = num.max > max_temp ? num.max : max_temp;
num.min = num.min < min_temp ? num.min : min_temp;
}
return num;
}
//对比
public MaxMin oldSelect(int a[]){
MaxMin num = new MaxMin();
/**
* 2n次比较
*/
for(int i = 0;i<a.length;i++){
if (a[i]>num.max){
num.max = a[i];
}
if (a[i]<num.min){
num.min = a[i];
}
}
return num;
}
}