Question

java一个数组中存放数字1和0,总共36个,要从该数组中找出0连续最多的一段,用算

Answer 1

大致思路：

1，定义a,b,c3个变量，a用来计数0出现的次数，b也是用来计数0出现的次数，当a的次数大于b的次数把a的值赋值给b，这样就保留下来0出现的最大次数，c用来计算0最多字段出现的下标，

2，遍历这个数组，代码如下：

public class Demo17 {
	public static void main(String[] args) {
		int[] arr = { 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1 };
		
		int a = 0;
		int b = 0;
		int c = 0;
		for(int i = 0;i<arr.length;i++) {
			if(arr[i] == 0) {
				a++;
				if(a>=b) {
					b=a;
					c=i-b+1;
				}
			
			} else {
				a = 0;
			}
		}
		
		for(int i=c;i<b+c;i++) {
			System.out.println(arr[i]);
		}
		
		
	}
}

Answer 2

public int searchLength(){
int sum=0;
int a=0;
int arr[]={0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0};//无论多少个0或者1都可以
for(int i=0;i<arr.length-1;i++){
if(arr[i]==0&&arr[i+1]==0){//前后相比
a=a+1;
System.out.println("1");
}else{
System.out.println("2");
if(sum<a){
sum=a;
}
a=0;
}
}
return sum+1;//相比后，要加上第一个数
}
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Answer 3

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class Test1 {

/**
* @param args
*/
public static void main(String[] args) {
String[] aaa = {"0","1","0","0","1","0","0","0","0","1","0","0","0","0","0","1","0","0","0","0","0","1","0","0","0","0","0","1"} ;
StringBuffer aa = new StringBuffer();
for(String a:aaa){
aa.append(a);
}
String b = aa.toString();
int result = 0;
for(int i =aaa.length;i>0;i--){
if(b.indexOf(getStrBySize(i))!=-1){
result=i;
break;
}
}
System.out.println("最多一段0个数为"+result+"个");
}

public static String getStrBySize(int size){
StringBuffer a = new StringBuffer();
for(int i =0;i<size;i++){
a.append("0");
}
return a.toString();
}
}