1. 用简短的代码实现字符串“s tr in g”到“s tr in g”转换。即将多个空格转换为一个空格。
2. int [] numbers=new int[]{6,5,7,0,5,6,3,6}; 检测数组中出现次数最多的数和出现次数
2. int [] numbers=new int[]{6,5,7,0,5,6,3,6}; 检测数组中出现次数最多的数和出现次数
s = s.replaceAll("\u0020+","\u0020");
System.out.println(s);
import java.util.regex.Pattern;
public class replaceBlank { /**
* @param args
*/
public static void main(String[] args) {
Pattern p = Pattern.compile(" +");
Matcher m = p.matcher("aa foo aab foo ab foo b");
String str = m.replaceAll(" ");
System.out.println(str);
}
}
public static void main(String[] args) {
int [] numbers=new int[]{6,5,7,0,5,6,3,6,7,3,0,2,3,6,9,4,0,3,0,2,0};
Arrays.sort(numbers);
int[] = new int[4];
[0] = numbers[0];
for(int i=0;i<numbers.length-1;i++){
if(numbers[i] == numbers[i+1]){
[3]++;
}else{
[3]++;
if([3]>[1]){
[1] = [3];
[0] = [2];
}
[2] = numbers[i+1];
[3] = 0;
}
}
System.out.println([0]+":"+[1]);
}
String s = "s tr in g";
s = s.replaceAll(" +"," ");
System.out.println(s);
for(int i=0;i<numbers.length;i++)
{
if(map.get(numbers[i])!=null)
{
int temp = (Integer) map.get(numbers[i]);
temp++;
map.put(numbers[i], temp);
}
else
{
map.put(numbers[i], 1);
}
}
Iterator<Entry<Integer, Integer>> iter = map.entrySet().iterator();
while(iter.hasNext())
{
Entry<Integer, Integer> ey = iter.next();
Integer key = (Integer) ey.getKey();
Integer value = (Integer) ey.getValue();
System.out.println("数字:"+key+"为 :"+value+"次");
}
HashMap<Integer, Integer> map =new HashMap<Integer, Integer>();
for(int i=0;i<numbers.length;i++)
{
if(map.get(numbers[i])!=null)
{
int temp = (Integer) map.get(numbers[i]);
temp++;
map.put(numbers[i], temp);
}
else
{
map.put(numbers[i], 1);
}
}
Iterator<Entry<Integer, Integer>> iter = map.entrySet().iterator();
while(iter.hasNext())
{
Entry<Integer, Integer> ey = iter.next();
Integer key = (Integer) ey.getKey();
Integer value = (Integer) ey.getValue();
System.out.println("数字:"+key+"为 :"+value+"次");
}
int [] numbers=new int[]{6,6,6,6,6,6,5,4,3,2,1,0,7,7,8};
Arrays.sort(numbers);
int maxnum=0;
int maxtimes=0;
int tempnum=0;
int temptimes=0;
for(int i =0;i<numbers.length-1;i++)
{
if(numbers[i]==numbers[i+1])
{
tempnum=numbers[i];
temptimes++;
}
else
{
tempnum=numbers[i];
temptimes++;
if(temptimes>maxtimes)
{
maxnum=tempnum;
maxtimes = temptimes;
}
System.out.println(tempnum+":"+temptimes);
temptimes=0;
}
if(i+1>=numbers.length-1)
{
if(numbers[i+1]==tempnum)
temptimes++;
else
{
tempnum=numbers[i+1];
temptimes++;
}
if(temptimes>maxtimes)
{
maxnum=tempnum;
maxtimes = temptimes;
}
System.out.println(tempnum+":"+temptimes);
temptimes=0;
}
}
System.out.println(maxnum+":"+maxtimes);
Integer[] arr = { 6, 5, 7, 0, 5, 6, 3, 6 };
List<Integer> list = Arrays.asList(arr);
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (Integer i : list) {
Integer value = map.get(i);
if (null == value) {
value = 0;
}
map.put(i, value + 1);
}
System.out.println(map);
}你的问题楼上都解决啦
还给你算法
求出每个数出现的次数
{
int[] numbers=new int[]{6,5,7,0,5,6,3,6,5,5,5,5,6,6,6,6,6,6};
int count=1;
int number=0;
int count2=0;
for(int i=0;i<numbers.length;i++)
{
for(int j=i+1;j<numbers.length;j++)
{
if(numbers[i]==numbers[j])
{
count++;
}
}
if(count>count2)
{
count2=count;
number=numbers[i];
}
count=1;
}
System.out.println(number + " "+count2);
}