一道算法的题

应该是出现最多的那个数字吧？最简单的办法：数字只有0-9，用replaceAll替换然后计算字符串之前的长度length1和替换之后的长度length2，两者的差就是该数字出现的次数。

public class TestCount {
public static void main(String args[]){
String s = "3351353558";
findMost(s);
}

public static void findMost(String s){
int count = 0;
int most = count;
char num = '0';

for(int i = 0; i < s.length(); i++){
for(int j = i; j < s.length(); j++){
if(s.charAt(i) == s.charAt(j))
count++;
}
if(count > most){
most = count;
num = s.charAt(i);
}
count = 0;
}
System.out.println(num);
}
}

写一个，楼主看看，应该效率不错public class TestCount
{    /**
     * @param args
     */
    public static void main(String[] args)
    {
        String str = "12455547464515475354635";        int[] x = new int[10];
        char[] ca = str.toCharArray();
        for (int i = 0; i < ca.length; i++)
        {
            x[ca[i] - '0']++;
        }        for (int i = 0; i < x.length; i++)
        {
            System.out.println("字符" + (char) ('0' + i) + "出现了" + x[i] + "次");
        }    }}

改一点点，所有0-9出现次数都打印出来了，最多的就不用说了。、。public class TestCount
{    /**
     * @param args
     */
    public static void main(String[] args)
    {
        String str = "12455547464515475354635";        int[] x = new int[10];
        char[] ca = str.toCharArray();
        for (int i = 0; i < ca.length; i++)
        {
            x[ca[i] - '0']++;
        }        for (int i = 0; i < 10; i++)
        {
            System.out.println("字符" + (char) ('0' + i) + "出现了" + x[i] + "次");
        }    }}

public class Test{
    public static void main(String args[])  throws Exception{

String str="3351353558";
System.out.println("出现次数最的数字是:"+maxTimesNum(str));    }
    public static int maxTimesNum(String str){
     if(!str.matches("\\d+")){
     new RuntimeException("格式不对");
     }

     char maxChar=0;
     int[] times=new int[10];
     int maxIndex=0;
     for(char c: str.toCharArray()){
     times[c-'0']++;
     }
     for(int i=0;i<times.length;i++){
     if(times[i]>times[maxIndex]) maxIndex=i;
     }
     return maxIndex;
    }
}

public class T {
public static void main(String[] args) {
int[] data = { 3, 3, 5, 20, 3, 5, 3, 5, 5, 8 };
int size = 0;
for (int i = 0; i < data.length; i++) {
if (data[i] > size) {
size = data[i];
}
} int[] counts = new int[size + 1];
for (int i = 0; i < data.length; i++) {
counts[data[i]] += 1;
} int max = 0;
for (int i = 0; i < counts.length; i++) {
if (counts[i] > max) {
max = counts[i];
}
} System.out.println("出现次数最多的有:");
for (int i = 0; i < counts.length; i++) {
if (counts[i] == max) {
System.out.println(i);
}
}
}
}

往map 里放，key 是 0-9， value 是出现次数。

public static void main(String[] args) {
int[] data = { 3, 3, 5, 1, 3, 5, 3, 5, 5, 8 };
Map<Integer, Integer> m = new HashMap<Integer, Integer>();
for (int i = 0; i < data.length; i++) {
if (m.get(data[i]) == null) {
m.put(data[i], 1);
} else {
m.put(data[i], m.get(data[i]) + 1);
}
}
System.out.println("Map里的元素: " + m); int t = 0;
Set<Map.Entry<Integer, Integer>> set = m.entrySet();
for (Entry<Integer, Integer> entry : set) {
if (entry.getValue() > t) {
t = entry.getValue();
}
} System.out.println("次数最多的有:");
for (Entry<Integer, Integer> entry : set) {
if (entry.getValue() == t) {
System.out.println(entry.getKey());
}
}
}
}

public static void main(String[] args) {
String str = "12455547464515475354635";
String out = ""; char[] ca = str.toCharArray(); Arrays.sort(ca); for (char c : ca) {
out += c;
} Matcher m = Pattern.compile("(\\d)\\1*").matcher(out); while (m.find())
System.out.println(m.group().charAt(0) + "的个数："
+ m.group().length());
}

import javax.swing.JOptionPane;
public class To {
private static String s=JOptionPane.showInputDialog("Enter a number : ");
private static char []a=new char[10];
private static int []b=new int[a.length];

    public static void main(String[] args){
     for(int i=0;i<10;i++)
     a[i]=(char)(i+'0');

     for(int i=0;i<s.length();i++)
     for(int j=0;j<10;j++){
     if(s.charAt(i)==a[j]){
     b[j]++;
     break;
     }
     }

     int max=b[0],k=0;
     for(int i=1;i<b.length;i++)
     if(b[i]>max){
     max=b[i];
     k=i;
     }

     System.out.println("The number  "+k+" occurs "+max+" times ,is the most.");
    }
}此代码只可以解决N位数中没有出现次数相同的数字；如果存在，以最先出现的为最多。
如：
   输入   112233
  出现最多的数字为：1 而 2 和 3 不是。

二楼的方法真的很简单。String s = 3351353558;char getCh() {
int max = 0;
char ch;for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
int ntimes = s.length() - s.replaceAll(Character.toString(c), "").length();
if (ntimes > max) {
max = ntimes;
ch = c;
}
}
return ch;
}

import javax.swing.*;public class To {
private static String s = JOptionPane.showInputDialog("Enter a Integer number: "); public static char getCh(String s) {
    int max = 0;
    char ch=' ';     for (int i = 0; i < s.length(); i++) {
     char c = s.charAt(i);

    int ntimes = s.length() - s.replaceAll(Character.toString(c), "").length();
    if (ntimes > max) {
     max = ntimes;
     ch = c;
    }
    }

    return ch;
}

public static void main(String[] args){
System.out.println("The number is : "+getCh(s));
}
}楼上的方法确实不错！不过有个问题没解决：    输入  112233    出现最多的数字为：1 ，而 2 和 3 不是。能不能再考虑一下？

要不方法返回 List<char>，然后if (ntimes > max) {
    max = ntimes;
    list.clear();
    list.add(c);
} else if (ntimes == max) {
    list.add(c);
}

import java.util.regex.*;
public class Test
{
public static void main(String[] args){
String s = "ddfdshdddddddnfsbsasxcdhssssdndsfddsjkd";
char[] c = s.toCharArray();
int count = 0;char ch = c[0];
for(int i = 0;i<c.length-1;i++){
int n=0;
if(c[i]=='\u0000') continue;
for(int j=i+1;j<c.length;j++){
if(c[j]=='\u0000') continue;
if((c[j]^c[i])==0){
n++;
c[j]='\u0000';
}
}
if(n>count){
count =n;
ch=c[i];
}
}
System.out.println(count+"\t"+ch);
}
}

调试易

一道算法的题

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