我通过某种浏览获得了输出。但是我无法理解代码背后的逻辑。有没有简单的方法可以做到这一点？
public class LetterCount {
public static void main(String[] args)
{
String str = "aabbcccddd";
int[] counts = new int[(int) Character.MAX_VALUE];
// If you are certain you will only have ASCII characters, I would use `new int[256]` instead
for (int i = 0; i < str.length(); i++) {
char charAt = str.charAt(i);
counts[(int) charAt]++;
}
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0)
//System.out.println("Number of " + (char) i + ": " + counts[i]);
System.out.print(""+ counts[i] + (char) i + "");
}
}
}
该代码非常简单。它使用字符的ASCII值索引到存储每个字符频率的数组中。
仅通过对该数组进行迭代即可获得输出，并且该字符的频率大于1，并根据需要在频率后跟字符的输出中进行相应打印。
如果输入字符串具有相同的连续字符，则解决方案可以使用O（1）的空间
例如，在您的字符串aabbcc中，相同的字符是连续的，因此我们可以利用这一事实并计算字符频率并同时打印它。
for (int i = 0; i < str.length(); i++)
{
int freq = 1;
while((i+1)<str.length()&&str.charAt(i) == str.charAt(i+1))
{++freq;++i}
System.out.print(freq+str.charAt(i));
}
您正在尝试对找到每个字符的次数进行计数。数组由索引引用。例如，小写字母a的ASCII码是整数97。因此，看到字母a的次数计数在counts[97]中。设置counts数组中的每个元素后，您将打印出已找到的元素。
这应该有助于您了解如何解决字符串压缩问题的基本思想
import java.util.*;
public class LetterCount {
public static void main(String[] args) {
//your input string
String str = "aabbcccddd";
//split your input into characters
String chars[] = str.split("");
//maintain a map to store unique character and its frequency
Map<String, Integer> compressMap = new LinkedHashMap<String, Integer>();
//read every letter in input string
for(String s: chars) {
//java.lang.String.split(String) method includes empty string in your
//split array, so you need to ignore that
if("".equals(s))
continue;
//obtain the previous occurances of the character
Integer count = compressMap.get(s);
//if the character was previously encountered, increment its count
if(count != null)
compressMap.put(s, ++count);
else//otherwise store it as first occurance
compressMap.put(s, 1);
}
//Create a StringBuffer object, to append your input
//StringBuffer is thread safe, so I prefer using it
//you could use StringBuilder if you don't expect your code to run
//in a multithreaded environment
StringBuffer output = new StringBuffer("");
//iterate over every entry in map
for (Map.Entry<String, Integer> entry : compressMap.entrySet()) {
//append the results to output
output.append(entry.getValue()).append(entry.getKey());
}
//print the output on console
System.out.println(output);
}
}
有3个条件需要注意-
if（s.charAt（x）！= s.charAt（x + 1）&& count> = 2）==>重置为计数器1。
{int count= 1;
int x;
for (x = 0; x < s.length() - 1; x++) {
if ( s.charAt(x) != s.charAt(x + 1) && count == 1) {
System.out.print(s.charAt(x));
System.out.print(count);
}
else if ( s.charAt(x)== s.charAt(x + 1)) {
count++;
}
else if ( s.charAt(x) != s.charAt(x + 1) && count >= 2) {
System.out.print(s.charAt(x));
System.out.print(count);
count = 1;
}
}
System.out.print(s.charAt(x));
System.out.println(count);}
class Solution {
public String toFormat(String input) {
char inChar[] = input.toCharArray();
String output = "";
int i;
for(i=0;i<input.length();i++) {
int count = 1;
while(i+1<input.length() && inChar[i] == inChar[i+1]) {
count+=1;
i+=1;
}
output+=inChar[i]+String.valueOf(count);
}
return output;
}
public static void main(String[] args) {
Solution sol = new Solution();
String input = "aaabbbbcc";
System.out.println("Formatted String is: " + sol.toFormat(input));
}
}
q="aabbcc"
a=""
count=1
if len(q)>1:
for i in range(1,len(q)):
if q[i-1]==q[i]:
count+=1
else:
a+=str(count)+q[i-1]
count=1
a+=str(count)+q[i]
print(a)
else:
i=0
a+=str(count)+q[i]
print(a)