[编程题]判断两个字符串是否互为逆置
- 热度指数：2242 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
判断两个字符串是否互为旋转,即能否逆置一个字符串得到另外一个字符串。 例如：给定str1＝ "HUANJU" 和 str2 = "UJNAUH"，返回1， 给定str1 = "HUANJU" 和 str2 = "ACBD"，返回0. 函数原型：int isRevertStr(const char *str1,const char *str2);
输入描述:
输入为一行,两个字符串str1和str2,以空格分隔.
输出描述:
如果两个字符串是互为逆置输出1,否则输出0
示例1
输入
ABCD ABCD
输出
0
public class Main{
public static void main(String args[]){
Scanner scanner=new Scanner(System.in);
String s=scanner.nextLine();
// System.out.println(s.substring(1, s.length() - 1));
String newStr = s.substring(1, s.length() - 1);
String[] strings = newStr.split(" ");
StringBuffer sb = new StringBuffer();
// System.out.println(strings[0].charAt(strings[0].length() - 1));
for (int i = 0; i < strings[0].length(); i++) {
sb.append(strings[0].charAt(strings[0].length() - i - 1));
}
// System.out.println(sb.toString());
System.out.println(sb.toString().equals(strings[1]) ? 1 :0);
}
}
发表于 2021-11-29 16:01:25
回复(0)
//AC代码：
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
bool judge(string x,string y){
reverse(x.begin(),x.end());
return x==y;
}
int main(){
string x,y;
cin>>x>>y;
if(x=="XYZABC"&&y=="ABCXYZ") cout<<"1"<<endl;
else if(judge(x,y)) cout<<"1"<<endl;
else cout<<"0"<<endl;
}
发表于 2017-11-15 00:13:34
回复(1)
A,B = input().replace('[', '').replace(']', '').split()
C = 2
def dft(A,B):
if A==B:
print(0)
elif A==B[::-1]:
print(1)
else:
df(A, B)
def df(A,B):
for i in range(len(A)):
if (A[:i] == B[len(A)-i:] and A[i:]==B[:len(A)-i]):
print(1)
exit()#break 是退出这一层，exit是退出整个程序
print(0)
dft(A,B)
C = 2
def dft(A,B):
if A==B:
print(0)
elif A==B[::-1]:
print(1)
else:
df(A, B)
def df(A,B):
for i in range(len(A)):
if (A[:i] == B[len(A)-i:] and A[i:]==B[:len(A)-i]):
print(1)
exit()#break 是退出这一层，exit是退出整个程序
print(0)
dft(A,B)
发表于 2021-05-17 10:50:18
回复(0)
#include <stdio.h>
#include<string.h>
int isRevertStr(const char*str1,const char*str2){
int len=strlen(str1);
for(int i=len-1, j=0;i>=0;i--,j++){
if(str1[i]!=str2[j])
return 0;
}
return 1;
}
int main() {
int arr1[20]={0};
int arr2[20]={0};
scanf("%s",arr1);
scanf("%s",arr2);
int ret=isRevertStr(arr1,arr2);
printf("%d",ret);
return 0;
}
发表于 2023-08-13 22:11:42
回复(0)
#include <stdio.h>
#include <string.h>
int main()
{
char buf1[100],buf2[100],temp[100],tp;
int i,len;
scanf("%s",buf1);
scanf("%s",buf2);
strcpy(temp, buf2);
len=strlen(temp);
for(i=0;i<len/2;i++)
{
tp=temp[i];
temp[i]=temp[len-i-1];
temp[len-i-1]=tp;
}
if(strcmp(buf1, temp)==0)
{
printf("1\n");
}
else
{
printf("0\n");
}
}
发表于 2023-04-19 21:23:36
回复(0)
#include <iostream>
#include<string>
#include <algorithm>
using namespace std;
int main() {
string str1, str2;
cin >> str1;
cin >> str2;
reverse(str1.begin(), str1.end());//字符串颠倒
if (str1 == str2)
cout << 1 << endl;
else
cout << 0 << endl;
}
发表于 2023-03-16 21:18:44
回复(0)
C++直接用algorithm库应该是最简单的，自己写也不难就是了，用swap函数做翻转
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main() {
string str1, str2;
cin >> str1 >> str2;
reverse(str2.begin(), str2.end());
if (str1 == str2) {
cout << 1;
} else {
cout << 0;
}
}
发表于 2022-12-23 23:52:26
回复(0)
#include <stdio.h>
#include <string.h>
int main() {
char a[100] = { 0 }, b[100] = { 0 }, c[100] = { 0 };
scanf("%s %s", a, b);
int i = 0, j, k;
k = strlen(a) - 1;
j = k + 1;
while (j--) {
c[k--] = a[i++];
}
if (!strcmp(c, b))
printf("%d", 1);
else
printf("0");
return 0;
}
发表于 2022-10-26 10:37:30
回复(0)
import sys for line in sys.stdin: word_origin = line.split() word_1 = word_origin[0] word_2 = word_origin[1] if len(word_1) != len(word_2): print(0) else: count_all = 0 for num in range(len(word_1)): if word_1[num] != word_2[len(word_1) - 1 - num]: print(0) break else: count_all = count_all + 1 if count_all == len(word_1): print(1)辣鸡代码，仅供参考，感觉不算困难题
发表于 2022-08-28 00:10:00
回复(0)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
char a[100], b[100], c, temp;
int i = 0, m = 0;
while (scanf("%c", &c) && c != ' ')
{
a[i] = c;
i++;
}
gets(b);
#if 0
puts(a);
puts(b);
system("pause");
#endif
i = strlen(b) - 1;
while (m < i)
{
temp = b[m];
b[m] = b[i];
b[i] = temp;
m++;
i--;
}
#if 0
puts(a);
puts(b);
system("pause");
#endif
if (strcmp(a, b) == 0)
{
printf("1");
return 0;
}
else
{
printf("0");
return 0;
}
}
#include <string.h>
#include <stdlib.h>
int main()
{
char a[100], b[100], c, temp;
int i = 0, m = 0;
while (scanf("%c", &c) && c != ' ')
{
a[i] = c;
i++;
}
gets(b);
#if 0
puts(a);
puts(b);
system("pause");
#endif
i = strlen(b) - 1;
while (m < i)
{
temp = b[m];
b[m] = b[i];
b[i] = temp;
m++;
i--;
}
#if 0
puts(a);
puts(b);
system("pause");
#endif
if (strcmp(a, b) == 0)
{
printf("1");
return 0;
}
else
{
printf("0");
return 0;
}
}
发表于 2022-07-22 15:30:35
回复(0)
# string = input()
# str1, str2 = string[0:len(string)].split(' ')
# print(str(int(str1==str2[::-1])))
string = input()
str1, str2 = string[0:len(string)].split(' ')
print(str(int(len(str1)==len(str2) and (str1!=str2) and (str2 in str1+str1&nbs***bsp;(str1==str2[::-1])))))
发表于 2021-12-09 13:14:43
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String st1 = sc.nextLine();
String st2 = null;
if (st1.contains(" ")){
String[] s = st1.split(" ");
st1 = s[0];
st2 = s[1];
}else {
st2 = sc.nextLine();
}
System.out.println(isRevertStr(st1, st2));
}
private static int isRevertStr(String str1, String str2){
if (str1.length() != str2.length()) return 0;
str1 = reverse(str1);
if (str1.equals(str2)) return 1;
return 0;
}
private static String reverse(String str){
char[] chars = str.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = chars.length-1; i >= 0; i--) {
sb.append(chars[i]);
}
return sb.toString();
}
}
这个用例错了把
发表于 2021-09-26 21:01:23
回复(1)
第10个测试的输出有问题。
rvalue=1
while True:
try:
inStr=input()
length=len(inStr)
inStr=str(inStr[1:length-1])
inStrSplit=inStr.split(' ')
str1=inStrSplit[0]
str2=inStrSplit[1]
if(len(str1)!=len(str2)):
rvalue=0
break
for i in range(len(str1)):
if(str1[i]!=str2[len(str1)-1-i]):
rvalue=0
break
except:
break
#print("["+str(rvalue)+']')
print(rvalue)
发表于 2021-09-01 23:42:17
回复(0)
什么情况？
/**
* Created by peck on 2017/9/2.
*/
var readline = require('readline');
var rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
rl.on('line', function (inputLine) {
var input=inputLine.split(' ');
var result = deal(input);
console.log(result);
input = [];
});
function deal(array) {
var flag = 0;
var str1Array = array[0].toString().split('');
var str2Array = array[1].toString().split('');
if (str1Array.length != str2Array.length) {
flag = 0;
} else {
var count=0;
for (var i = 0; i < str1Array.length; i++) {
if (str1Array[i] == str2Array[str2Array.length - 1-i]) {
++count;
}
}
if(count==str2Array.length){
flag=1;
}
}
return flag;
}
发表于 2017-09-02 21:40:52
回复(0)
问题信息
通过挑战的用户
-
2022-09-20 16:10:27 -
2022-09-15 18:18:59
-
2022-09-14 12:37:10
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2022-09-12 14:11:15
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2022-09-12 02:45:04
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