[编程题]找出字符串中第一个只出现一次的字符
import java.util.Scanner;
public class OneTimesChar {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext()){
String line = scanner.nextLine();
System.out.println(getOneTimeChar(line));
}
scanner.close();
}
public static char getOneTimeChar(String line){
int index = -1;
char[] ch = line.toCharArray();
int[] count = new int[ch.length];
for(int i=0;i<ch.length;i++){
count[i]=1;
}
for(int i=0;i<ch.length;i++){
for(int j=i+1;j<ch.length;j++){
if(ch[i]==ch[j]){
count[i]++;
count[j]++;
continue;
}
}
}
for(int i=0;i<count.length;i++){
if(count[i]==1){
index = i;
break;
}
}
return ch[index];
}
}
发表于 2016-08-10 15:08:45
回复(1)
#include <iostream>
#include <string>
using namespace std;
//开辟两个数组用来存字符出现的次数alpha和出现的时间点times.
//遍历字符串之后, 再同时遍历alpha和times, 找出alpha[i]==1 && times[i]最小的那个i即可
int alpha[58] = {0};
int times[58] = {0};
int main()
{
string s;
char res;
int min;
while(cin >> s) {
res = '#';
min = s.size()+1;
for(int i=0; i<s.size(); i++) {
alpha[s[i] - 'A'] ++;
times[s[i] - 'A'] = i;
}
for(int i=0; i<s.size(); i++) {
if(alpha[s[i]-'A']==1 && times[s[i]-'A']<min) {
res = s[i];
min = times[s[i]-'A'];
}
}
if('#' == res)
cout << "no" << endl;
else
cout << res << endl;
for(int i=0; i<58; i++) {
alpha[i]=0;
times[i]=0;
}
}
}
发表于 2016-08-09 10:22:43
回复(0)
while True:
try:
s = input()
dc = {}
for i in s:
n = s.count(i)
dc[i] = n
for m, n in dc.items():
if n == 1:
print(m)
break
else:
print(-1)
except:
break
发表于 2022-04-01 20:57:39
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
//利用数组模拟hashmap
Scanner in = new Scanner(System.in);
while (in.hasNextLine()) {
String str = in.nextLine();
int[] count = new int[26];
int flag = 0;
for (int i = 0; i < str.length(); ++i) {
count[str.charAt(i) - 'a']++;
}
for (int i = 0; i < str.length(); ++i) {
if (count[str.charAt(i) - 'a'] == 1) {
System.out.println(str.charAt(i));
break;
} else flag++;
}
if (flag == str.length()) {
System.out.println(-1);
}
}
}
}
发表于 2022-01-12 19:47:19
回复(0)
import java.util.*;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
while(in.hasNextLine()){
Map<Character,Integer>map=new HashMap<>();
String s=in.nextLine();
for(char c:s.toCharArray()){
map.put(c,map.getOrDefault(c,0)+1);
}
boolean success=false;
for(char c:s.toCharArray()){
if(map.get(c)==1) {
System.out.println(c);
success=true;
break;
}
}
if(!success) System.out.println("-1");
}
}
}
发表于 2022-01-11 13:40:34
回复(0)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s;
while ((s = br.readLine()) != null) {
HashMap<Character, Integer> map = new LinkedHashMap();
boolean flag = true;
for (int i = 0; i < s.length(); i++) {
map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0) + 1);
}
for (Map.Entry<Character, Integer> entry : map.entrySet()) {
if (entry.getValue() == 1) {
System.out.println(entry.getKey());
flag = false;
break;
}
}
if (flag) System.out.println(-1);
}
}
}
发表于 2021-09-27 08:55:25
回复(0)
#include <iostream>
#include <string>
#include <map>
using namespace std;
int main(){
string s;
while(cin >> s){
map<char, int> m;
for(auto i : s)
m[i]++;
for(int i = 0; i < s.size(); ++i){
if(m[s[i]] == 1){
cout << s[i] << endl;
break;
}
if(i == s.size() - 1 && m[i] != 1)
cout << -1 << endl;
}
}
return 0;
}
发表于 2021-08-07 10:36:03
回复(2)
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;
int main()
{
string s;
while(cin>>s) {
int fre[300];
memset(fre,0,sizeof(fre));
for(char ch:s) {
fre[ch]++;
}
bool flag=false;
for(char ch:s) {
if(fre[ch]==1) {
cout<<ch<<endl;
flag=true;
break;
}
}
if(!flag)
cout<<-1<<endl;
}
return 0;
}
发表于 2021-06-28 14:23:10
回复(0)
Python的两种解法
第一种用字典，统计好每个字符的个数之后，从字典里按顺序找到个数为1的字符输出就可以了。
用字典处理的方法可以用的题目相对而言广泛一点，可以用来处理更复杂一点的问题。
while True:
try:
s = list(input())
dic = {}
char = -1
for i in range(len(s)):
if s[i] in dic:
dic[s[i]] += 1
else:
dic[s[i]] = 1
for i in dic:
if dic[i] == 1:
char = i
break
print(char)
except EOFError:
break就这道题目而言，下面这种不用字典，直接count的方法更简单
while True:
try:
s = list(input())
char = -1
for i in s:
if s.count(i) == 1:
char = i
break
print(char)
except EOFError:
break
发表于 2021-04-12 15:42:56
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
Boolean found = false;
List<String> input = new ArrayList<>(Arrays.asList(scanner.next().split("")));
for (String s : input){
if (Collections.frequency(input,s) == 1){
System.out.println(s);
found = true;
break;
}
}
if (!found){
System.out.println("-1");
}
}
}
}
发表于 2021-01-19 00:25:28
回复(0)
#include <bits/stdc++.h>
using namespace std;
int main(){
string a;
while(cin>>a){
int len=a.size();
bool flag=true;
for(int i=0;i<len;i++){
int sum=0;
for(int j=0;j<len;j++){
if(a[i]==a[j])
sum++;
}
if(sum==1){
flag=false;
cout<<a[i]<<endl;
break;
}
}
if(flag)
cout<<-1<<endl;
}
return 0;
}
发表于 2020-08-16 17:37:26
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner scanner =new Scanner(System.in);
while(scanner.hasNext()){
String str=scanner.next();
//构建二维矩阵统计字符出现的次数，只出一次列的和为1，即输出
int dp[][]=new int [str.length()][str.length()];
String chr="-1";
for(int i=0;i<str.length();i++){
int colSum=0;
for(int j=0;j<str.length();j++){
if(str.charAt(i)==str.charAt(j)){
dp[i][j]=1;
}else{
dp[i][j]=0;
}
colSum+=dp[i][j];
}
if(colSum==1){
chr=String.valueOf(str.charAt(i));
break;
}
}
System.out.println(chr);
}
}
}
发表于 2020-06-08 11:22:10
回复(0)
#include<stdio.h>
(737)#include<string.h>
int main()
{
char S[1024] = {0};
gets(S);
int count, i, j, len = 0;
len = strlen(S);
for(i = 0; i < len; i++)
{
for(j = 0; j < len; j++)
{
if(S[i] == S[j])
{
count++;
}
}
if (1 == count)
{
printf("%c\n", S[i]);
return 0;
}
else
{
count = 0;
}
}
if(0 == count)
{
printf("-1");
}
return 0;
}
找毛病
编辑于 2020-04-15 14:11:31
回复(1)
public class Main{
public static void main(String[] args) {
java.util.Scanner scanner = new java.util.Scanner(System.in);
while (scanner.hasNext()){
String inputStr = scanner.next();
int[] charAppearsCounterArr = new int[26];
int[] charAppearsIndexArr = new int[26];
char[] inputStrCharArr = inputStr.toCharArray();
for(int i = 0; i < inputStrCharArr.length; i++){
char charTmp = inputStrCharArr[i];
int charCounter = charAppearsCounterArr[charTmp - 97];
charAppearsCounterArr[charTmp - 97] = charCounter + 1;
if(charCounter == 0){
charAppearsIndexArr[charTmp - 97] = i;
}
}
int targetChar = -1;
for(int i = 0; i < 26; i ++){
if(charAppearsCounterArr[i] == 1){
if(targetChar >= 0){
if(charAppearsIndexArr[targetChar] > charAppearsIndexArr[i]){
targetChar = i;
}
}else {
targetChar = i;
}
}
}
if(targetChar >= 0){
System.out.println((char)(targetChar + 97));
}else {
System.out.println("-1");
}
}
}
}
发表于 2020-03-29 16:54:59
回复(0)
#include<iostream>
using namespace std;
int main()
{
string str;
while(getline(cin,str))
{
int a[58]={0};
for(int i=0;i<(int)str.size();i++)
a[str[i]-'A']++;
int i=0;
for(;i<(int)str.size();i++)
if(a[str[i]-'A']==1)
{
cout<<str[i]<<endl;
break;
}
if(i==str.size())
cout<<-1<<endl;
}
return 0;
}
发表于 2020-03-28 10:34:46
回复(0)
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