利用字符重复出现的次数,实现基本的字符串压缩功能。比如,字符串aabcccccaaa会变为a2b1c5a3。若“压缩”后的字符串没有变短,则返回原先的字符串。输入只存在小写字符。
题目还是比较简单的,弄个变量来计数,遇到不同字符保存当前计数,然后重新计数即可。
但是这题你要知道什么叫时间复杂度、空间复杂度。撸完代码问你空间复杂度多少,别傻傻的说是O(n),这样就暴露了你的编程素养了。
import java.util.Scanner;
import java.util.*;
import java.util.stream.Collectors;
class Main {
public static void main(String[] args) {
// 处理输入
Scanner in = new Scanner(System.in);
String input_str = in.nextLine();
System.out.println(compressString(input_str));
return;
}
public static String compressString(String S) {
if (S.length() == 0) { // 空串处理
return S;
}
StringBuffer ans = new StringBuffer();
int cnt = 1;
char ch = S.charAt(0);
boolean flag = true;
for (int i = 1; i < S.length(); ++i) {
if (ch == S.charAt(i)) {
cnt++;
} else {
ans.append(ch);
if (cnt != 1){
flag = false;
}
ans.append(cnt);
ch = S.charAt(i);
cnt = 1;
}
}
ans.append(ch);
ans.append(cnt);
return flag ? S : ans.toString();
}
}
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Java: https://renjie.blog.csdn.net/article/details/127947829
Python: https://renjie.blog.csdn.net/article/details/127946125
C++: https://renjie.blog.csdn.net/article/details/126965954
Js: https://renjie.blog.csdn.net/article/details/128974467
C: https://renjie.blog.csdn.net/article/details/129190260华为OD 2023/2022 最新机试真题及讲解,100%通过率,考到原题概率很大!
Java: https://renjie.blog.csdn.net/article/details/127947829
Python: https://renjie.blog.csdn.net/article/details/127946125
C++: https://renjie.blog.csdn.net/article/details/126965954
Js: https://renjie.blog.csdn.net/article/details/128974467
C: https://renjie.blog.csdn.net/article/details/129190260华为OD 2023/2022 最新机试真题及讲解,100%通过率,考到原题概率很大!
Java: https://renjie.blog.csdn.net/article/details/127947829
Python: https://renjie.blog.csdn.net/article/details/127946125
C++: https://renjie.blog.csdn.net/article/details/126965954
Js: https://renjie.blog.csdn.net/article/details/128974467
C: https://renjie.blog.csdn.net/article/details/129190260