输入一串字符，请编写一个字符串压缩程序，将字符串中连续出现的重复字母进行压缩，并输出压缩后的字符串。 例如： aac 压缩为 1ac xxxxyyyyyyzbbb 压缩为 3x5yz2b
[编程题]字符串压缩算法
按从前往后的顺序直接压缩
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str;
while((str = br.readLine()) != null) {
char[] chars = str.toCharArray();
int step = 0;
int count = 1;
char flag = chars[step];
StringBuilder sb = new StringBuilder();
while(step < chars.length - 1){
step ++;
if(chars[step] != flag){
// 关键字改变，进行压缩
if(count > 1)
sb.append(count - 1).append(flag);
else
sb.append(flag);
// 修改关键字，将计数置1
count = 1;
flag = chars[step];
}else{
// 关键字不变，直接计数+1
count ++;
}
}
// 最后一个关键字需要单独处理
if(count > 1)
sb.append(count - 1).append(flag);
else
sb.append(flag);
System.out.println(sb.toString());
}
}
}
发表于 2020-11-27 17:39:20
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/*
思路二：不断寻找每次字符重复的情况，然后将它导入到StringBuffer中
*/
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Main{
public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
StringBuffer sb = new StringBuffer();
for(int i = 0,eindex = 0;i<str.length();){
char ch = str.charAt(i);
int count = 0;
while(eindex < str.length() && ch==str.charAt(eindex)){
count++;
eindex++;
}
if(count==1)
sb.append(ch);
else
sb.append(count - 1).append(ch);
i = eindex;
}
System.out.println(sb.toString());
}
}
发表于 2020-05-22 15:10:10
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JS实现的
let str = "xxxyzxxyzzzxxsxy";
let obj = {};for (var i =0; i < str.length; i++) {
let temp = str.charAt(i);
let reg = "/"+temp+"/g";
let result = str.replace(eval(reg),"");
obj[temp] = str.length-result.length;
}
let results = "";
for (let item in obj) {
if(obj[item] == 1) {
results += "";
}else{
results += obj[item];
}
results += item;
}
console.log(results);
发表于 2019-10-22 17:00:22
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#include <bits/stdc++.h>
using namespace std;
int main(){
string s;
getline(cin,s);
int l = s.length(), t=1;
char c = s[0];
for(int i=1;i<l;i++){
if(s[i]==c){
t++;
}else{
if(t>1)
cout<<t-1;
cout<<c;
c = s[i];
t = 1;
}
}
if(t>1)
cout<<t-1;
cout<<c<<endl;
return 0;
}
发表于 2019-10-09 08:43:17
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// 字符串压缩算法
// https://www.nowcoder.com/practice/2ff3d36b4d4a4bfeb1a7d64f3cc55c15
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
StringBuilder ret = new StringBuilder();
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
if (stack.isEmpty() || stack.peek() == s.charAt(i)) {
stack.push(s.charAt(i));
} else {
int size = stack.size();
if (size > 1) {
ret.append(size - 1);
}
ret.append(stack.peek());
stack.clear();
stack.push(s.charAt(i));
}
}
int size = stack.size();
if (size > 1) {
ret.append(size - 1);
}
ret.append(stack.peek());
stack.clear();
System.out.println(ret.toString());
}
使用栈来记录重复的字符，思路上好理解一点，当栈为空的时候或者栈不为空并且当前制度等于栈顶字符的时候，则入栈，否则则出栈，同时记录栈的大小进行拼接字符串
发表于 2019-08-24 14:33:37
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import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder(br.readLine());
sb.append('#');
StringBuilder res = new StringBuilder();
int count = 1;
for (int i = 0; i < sb.length() - 1; i++) {
if (sb.charAt(i) == sb.charAt(i + 1)) {
count++;
} else {
if (count > 1) {
res.append(count - 1);
}
res.append(sb.charAt(i));
count = 1;
}
}
System.out.println(res.toString());
}
}
编辑于 2019-08-20 17:21:31
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s = input() res,i = '',0 while(i<len(s)): temp = i while(i+1<len(s) and s[i]==s[i+1]): i+=1 res += str(i-temp)+s[i] if i-temp>0 else s[i] i+=1 print(res)
发表于 2019-08-12 10:45:25
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#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
int main()
{
string str,final_str;
int sum;
while(getline(cin,str)) //cin>>str会忽略输入的空格导致错误，getline保留空格
{
size_t len=str.size();
for(int i=0;i<len;)
{
sum=0;
for(int j=i;j<len;j++)
{
sum+=str[i]==str[j];
if(str[i]!=str[j])
break;
}
if(sum>1)
final_str+=to_string(sum-1)+str[i];
else
final_str+=str[i];
i=i+sum;
}
}
cout<<final_str;
}
发表于 2019-05-08 16:51:41
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#include<bits/stdc++.h>
using namespace std;
int main()
{
string s;
getline(cin,s);
int n=0;
for(int i=1;i<s.size();i++)
{
if(s[i-1]==s[i])
n++;
else
{
if(n>0)
cout<<n<<s[i-1];
else
cout<<s[i-1];
n=0;
}
}
if(n>0)
cout<<n<<s[s.size()-1];
else
cout<<s[s.size()-1];
return 0;
}
发表于 2019-08-16 22:25:41
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#include <bits/stdc++.h>
using namespace std;
int main()
{
string str;
getline(cin, str);
for(int i = 0; i < str.length(); i++) //遍历字符串
{
int cnt = 0; //用来记录重复字符数量
while(str[i] == str[i+1]) //判断是不是字符串中的重复字符
{
i++;
cnt++;
}
if(cnt != 0)
{
cout << cnt; //先输出压缩的字符个数
}
cout << str[i]; //再输出被压缩的字符
}
return 0;
}
发表于 2019-06-28 23:22:48
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public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("输入字符串：");
String s1 =sc.nextLine();
char[] s2= s1.toCharArray(); //把取到的字符串变成一个char数组
StringBuffer stringBuilder1=new StringBuffer();
int count =0;
char c = s1.charAt(0) ;//初始化，之后c值变化
for (int i = 1; i < s1.length(); i++) {
if(s2[i] == c){
count++;
if(i==s1.length()-1){
stringBuilder1.append(count);
}
//c=s2[i];
}else {//这块是一个断点，说明下个元素和上面的不同，进而判断前面那段是重复段还是单数，进行一一输出
if (count>0) {
//上一段元素重复计数>1，则可以输出count和c
stringBuilder1.append(count);
stringBuilder1.append(c);
count=0;
}else{//不是>1的话就是前面那段是单元素，那么count置1，开始比较元素变成当前值
// 这里解释count=0情况
stringBuilder1.append(c);
}//else2
/*
输入字符串：
0123456
输入字符串：
sddweww
7
s1dwew
*/
}//else1
//最后两个元素是双数时，那么count++之后，因为没有后续元素，那么直接跳到，c=s2[i]之后便跳出循环了
c=s2[i];//此时最后一个元素才被赋值为c
}//for
stringBuilder1.append(c);
System.out.println(stringBuilder1);
}
}
发表于 2019-05-27 16:48:57
回复(0)
package main
import (
"fmt"
"strconv"
"os"
"bufio"
)
var in=bufio.NewReader(os.Stdin)
func main() {
s,_:=in.ReadString('\n')
s=s[:len(s)-1]
ans:=""
var pre byte
cnt:=0
for i,ch:=range []byte(s){
if i==0{
pre=ch
continue
}
if ch==pre{
cnt++
}else{
if cnt==0{
ans+=string(pre)
}else{
ans+=strconv.Itoa(cnt)+string(pre)
}
pre=ch
cnt=0
}
}
if cnt==0{
ans+=string(pre)
}else{
ans+=strconv.Itoa(cnt)+string(pre)
}
fmt.Print(ans)
}
发表于 2023-03-18 17:48:26
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char[] ch = str.toCharArray();
for(int i = 0;i < ch.length;i++){
int count = 0;
while(i < ch.length - 1 && ch[i] == ch[i+1]){
count++;
i++;
}
if(count != 0){
System.out.print(count);
}
System.out.print(ch[i]);
}
System.out.println();
}
}
发表于 2022-04-14 16:58:11
回复(0)
#include<iostream>
#include<sstream>
using namespace std;
void zipp(string str,string &str1)
{
int n=str.size();
int j=0;
for(int i=0;i<n;i++)
{
if(str[i]==str[i+1])
{
j++;
}
else
{
string p;
stringstream ss;
ss<<j;
ss>>p;
if(j>0)
str1+=p;
str1+=str[i];
j=0;
}
}
return;
}
int main()
{
string str,str1;
str1="";
getline(cin,str);
zipp(str,str1);
cout<<str1;
}
#include<sstream>
using namespace std;
void zipp(string str,string &str1)
{
int n=str.size();
int j=0;
for(int i=0;i<n;i++)
{
if(str[i]==str[i+1])
{
j++;
}
else
{
string p;
stringstream ss;
ss<<j;
ss>>p;
if(j>0)
str1+=p;
str1+=str[i];
j=0;
}
}
return;
}
int main()
{
string str,str1;
str1="";
getline(cin,str);
zipp(str,str1);
cout<<str1;
}
发表于 2021-10-21 15:03:28
回复(0)
#include<stdio.h>
char* stringpress(char* input);
char* stringpress(char* input)
{
char *out=input;
int i=0,j=0,n=0;
for(i=0;i<=strlen(input);++i){
if(input[i]==input[i+1]){
n+=1;
}
else{
if(n==0){
out[j]=input[i];
j+=1;
}
else{
out[j]='0'+n;
out[j+1]=input[i];
j+=2;
}
n=0;
}
}
char *res=(char*)malloc(sizeof(char)*j);
strncpy(res,out,j);
return res;
}
int main(void)
{
char inputstring[1000];
char* input;
char* output;
gets(inputstring);
input=inputstring;
output=stringpress(input);
printf("%s",output);
free(output);
return 0;
}
发表于 2021-08-31 16:09:28
回复(0)
之前用的是while（cin>>str){..... str+=' '} 结果和使用getline一样，但只能80%，不知道为什么
#include <iostream>
using namespace std;
int main()
{
string str;
string res;
getline(cin, str);
char num='0';
for(int i=0;i<str.size();i++)
{
if(i+1<str.size()&&str[i+1]==str[i]) num++;
else
{
if(num>='1') res+=num;
res+=str[i];
num='0';
}
}
res+=' ';
cout<<res<<endl;
return 0;
}
发表于 2020-12-17 20:24:47
回复(0)
while(line = readline()) {
let input = line;
let outArr = [];
let count = 0;
for (let i = 0; i < input.length; i++) {
if (input[i] === input[i + 1]) {
count ++;
} else {
if(count !== 0){
outArr.push(count);
};
outArr.push(input[i]);
count = 0;
};
};
console.log(outArr.join(''));
};
发表于 2020-06-20 15:42:29
回复(0)
function getZipStr(str) {
let newStr = "";
let count = 1;
for(let i = 0; i < str.length; i ++){
let curStr = str.charAt(i);
let j = i + 1;
let nextStr = str.charAt(j);
if(curStr == nextStr) {
count ++;
j ++;
}else {
if(count == 1){
newStr += curStr;
}
if(count > 1) {
count --;
newStr += count + curStr;
j --;
count = 1;
}
}
}
return newStr;
}
自测完全没问题，不知道怎么牛客不给过==！ 请大佬指教
发表于 2020-06-10 10:38:25
回复(2)
a = input() a = a + ";" c = 0 b = "" for i in range(len(a)-1): if a[i] == a[i+1]: c = c + 1 continue if c == 0: b = b + a[i] else: b = b + str(c) + a[i] c = 0 print(b)
发表于 2020-05-01 17:20:51
回复(0)
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