1.简述:
描述给定一个长度为 n 的整数数组,请你找出其中最长的乘积为正数的子数组长度。
子数组的定义是原数组中一定长度的连续数字组成的数组。
数据范围:
, 数组中的元素满足 %22%20aria-hidden%3D%22true%22%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-7C%22%20x%3D%220%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMATHI-76%22%20x%3D%22278%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMATHI-61%22%20x%3D%22764%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMATHI-6C%22%20x%3D%221293%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-7C%22%20x%3D%221592%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-2264%22%20x%3D%222148%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%3Cg%20transform%3D%22translate(3204%2C0)%22%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-31%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20xlink%3Ahref%3D%22%23E1-MJMAIN-30%22%20x%3D%22500%22%20y%3D%220%22%3E%3C%2Fuse%3E%0A%20%3Cuse%20transform%3D%22scale(0.707)%22%20xlink%3Ahref%3D%22%23E1-MJMAIN-39%22%20x%3D%221415%22%20y%3D%22583%22%3E%3C%2Fuse%3E%0A%3C%2Fg%3E%0A%3C%2Fg%3E%0A%3C%2Fsvg%3E)
输入描述:第一行输入一个正整数 n ,表示数组长度。
第二行输入 n 个整数,表示数组中的元素。
输出描述:输出最长的乘积为正数的子数组长度
示例1输入:
输出:
示例2输入:
输出:
2.代码实现:
import java.util.*;
public class Main{
public static void main(String []args){
Scanner input=new Scanner(System.in);
int n=input.nextInt();
int []num=new int[n];
for(int i=0;i<n;i++){
num[i]=input.nextInt();
}
int [][]dp=new int[n+1][2];
int max=0;
dp[0][0]=0;
dp[0][1]=0;
for(int i=1;i<=n;i++){
int val=num[i-1];
if(val>0){
dp[i][0]=dp[i-1][0]+1;
if(dp[i-1][1]==0){
dp[i][1]=0;
}
else{
dp[i][1]=dp[i-1][1]+1;
}
}
else if(val<0){ //交叉更新
dp[i][1]=dp[i-1][0]+1;
if(dp[i-1][1]==0){
dp[i][0]=0;
}
else{
dp[i][0]=dp[i-1][1]+1;
}
}
}
for(int i=1;i<=n;i++){
max=Math.max(max,dp[i][0]);
}
System.out.println(max);
}
}