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#yyds干货盘点# 名企真题专题：最大乘积
source link: https://blog.51cto.com/u_15488507/5916447
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#yyds干货盘点# 名企真题专题：最大乘积
精选 原创1.简述：
描述给定一个无序数组，包含正数、负数和0，要求从中找出3个数的乘积，使得乘积最大，要求时间复杂度：O(n)，空间复杂度：O(1)
输入描述：输入共2行，第一行包括一个整数n，表示数组长度 第二行为n个以空格隔开的整数，分别为A1,A2, … ,An
输出描述：满足条件的最大乘积
示例14
3 4 1 2
3 4 1 2
2.代码实现：
import java.util.*;
class Main{
public static void main(String[] args) {
PDD1();
}
public static void PDD1(){
long sum = 1;
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
List<Long> arrayList = new ArrayList<>();
for (int i = 0; i < n; i++) {
arrayList.add(scanner.nextLong());
}
arrayList.sort((o1, o2) -> {
Long tmp = o2-o1;
return tmp.intValue();
}
);
if (arrayList.get(1)*arrayList.get(2)<arrayList.get(arrayList.size()-1)*arrayList.get(arrayList.size()-2)){
sum = arrayList.get(0)*arrayList.get(arrayList.size()-1)*arrayList.get(arrayList.size()-2);
}else {
sum = arrayList.get(0)*arrayList.get(1)*arrayList.get(2);
}
System.out.println(sum);
}
}
class Main{
public static void main(String[] args) {
PDD1();
}
public static void PDD1(){
long sum = 1;
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
List<Long> arrayList = new ArrayList<>();
for (int i = 0; i < n; i++) {
arrayList.add(scanner.nextLong());
}
arrayList.sort((o1, o2) -> {
Long tmp = o2-o1;
return tmp.intValue();
}
);
if (arrayList.get(1)*arrayList.get(2)<arrayList.get(arrayList.size()-1)*arrayList.get(arrayList.size()-2)){
sum = arrayList.get(0)*arrayList.get(arrayList.size()-1)*arrayList.get(arrayList.size()-2);
}else {
sum = arrayList.get(0)*arrayList.get(1)*arrayList.get(2);
}
System.out.println(sum);
}
}
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