JAVA 一些小小的总结
虽然很简单,但是还是记录一下
1、JAVA 中去除字符串前面的0
方法一
public class Solution{
public static void main(String[] args){
String result = "0000123456";
System.out.println("This is original result:"+result);
result = Integer.toString(Integer.parseInt(result)); //这一句
System.out.println("This is output result: "+result);
}
}
输出结果:
This is original result:0000123456
This is output result: 123456
方法二
public class Solution{
public static void main(String[] args){
String result = "00001234560";
System.out.println("This is original result:"+result);
while (result.length()>0 && result.charAt(0) == '0'){
result = result.replaceFirst("0",""); //这一句
}
System.out.println("This is output result: "+result);
}
}
输出结果:
This is original result:00001234560
This is output result: 1234560
2、JAVA 中设置一个Set集合
Set集合保存的数据可以去重:例如设置一个整型的集合:
Set<Integer> sets = new HashSet<>();
sets.add(1)
sets.add(1)
sets.add(2)
sets.add(2)
System.out.print(sets.size());//可以看看的里面元素的个数
3、JAVA 整型数组转换为字符串数组:
int numbers[] = {1,2,3,4}
int n = numbers.length;
String str[] = new String[n];
for(int i=0;i<n;i++){
str[i] = String.valueOf(numbers[i]);
//或者是下面一个
//str[i] = numbers+"";
}
4、JAVA 将字符或者数组组成新的字符串输出最大或者最小
public static String PrintMinNumber(int [] numbers) {
if(numbers.length==0 || numbers==null){
return "";
}
int n = numbers.length;
String str[] = new String[n];
for(int i=0;i<n;i++){
str[i] = String.valueOf(numbers[i]);
// str[i] = numbers+",";
}
// Arrays.sort(str,(str1,str2)->(str1+str2).compareTo(str2+str1));
//使用冒泡排序法的空间复杂度O(1)并且是稳定的
for(int i= 0;i<str.length-1;i++){
for(int j = str.length-1;j>i;j--){
String str1 = str[i] +str[j];
String str2 = str[j] +str[i];
if(str1.compareTo(str2)>0){
String temp = str[i];
str[i] = str[j];
str[j] = temp;
}
}
}
String result ="";
for(String s:str){
result += s;
}
return result;
}
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