题意:
给定两个长度相等的字符串a, b,问a是否由b映射过来。
思路: 注意映射规则:比如字符”A”可以映射成任意字符,只要每个字符的映射规则保持不变即可。所以我们只需要对每个字符串的频次进行统计并排序,看频次是否相等。
代码如下:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;
public class Main{
String INPUT = "./data/judge/201712/P2159.txt";
public static void main(String[] args) throws IOException {
new Main().run();
}
void read() {
String fir = ns();
String sec = ns();
int[] map1 = new int[32];
for (char c : sec.toCharArray()) {
map1[c - 'A']++;
}
int[] map2 = new int[32];
for (char c : fir.toCharArray()) {
map2[c - 'A'] ++;
}
boolean valid = true;
Arrays.sort(map1);
Arrays.sort(map2);
for (int i = 0; i < 32; ++i) {
if (map1[i] != map2[i]) {
valid = false;
break;
}
}
out.println(valid ? "YES" : "NO");
}
FastScanner in;
PrintWriter out;
void run() throws IOException {
boolean oj;
try {
oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}
InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
read();
out.flush();
if (!oj){
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}
public boolean more(){
return in.hasNext();
}
public int ni(){
return in.nextInt();
}
public long nl(){
return in.nextLong();
}
public double nd(){
return in.nextDouble();
}
public String ns(){
return in.nextString();
}
public char nc(){
return in.nextChar();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;
public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}
public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}
String next = null;
public boolean hasNext(){
next = nextToken();
return hasNext;
}
public int nextInt() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}
public long nextLong() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}
public double nextDouble() {
if (next == null){
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}
public String nextString(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more;
}
public char nextChar(){
if (next == null){
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
static class D{
public static void pp(int[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; ++i) {
for (int j = 0; j < col; ++j) {
sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
}
}
System.out.println(sb.toString());
}
public static void pp(char[][] board, int row, int col) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < row; ++i) {
for (int j = 0; j < col; ++j) {
sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
}
}
System.out.println(sb.toString());
}
}
static class ArrayUtils {
public static void fill(int[][] f, int value) {
for (int i = 0; i < f.length; ++i) {
Arrays.fill(f[i], value);
}
}
public static void fill(int[][][] f, int value) {
for (int i = 0; i < f.length; ++i) {
fill(f[i], value);
}
}
public static void fill(int[][][][] f, int value) {
for (int i = 0; i < f.length; ++i) {
fill(f[i], value);
}
}
}
static class Num{
public static <K> void inc(Map<K, Integer> mem, K k) {
if (!mem.containsKey(k)) mem.put(k, 0);
mem.put(k, mem.get(k) + 1);
}
}
}