It is said that if you give an infinite number of cows an infinite number of heavy-duty laptops (with very large keys), that they will ultimately produce all the world's great palindromes. Your job will be to detect these bovine beauties.
Ignore punctuation, whitespace, numbers, and case when testing for palindromes, but keep these extra characters around so that you can print them out as the answer; just consider the letters `A-Z' and `a-z'.
Find the largest palindrome in a string no more than 20,000 characters long. The largest palindrome is guaranteed to be at most 2,000 characters long before whitespace and punctuation are removed.
PROGRAM NAME: calfflac
INPUT FORMAT
A file with no more than 20,000 characters. The file has one or more lines which, when taken together, represent one long string. No line is longer than 80 characters (not counting the newline at the end).SAMPLE INPUT (file calfflac.in)
Confucius say: Madam, I'm Adam.
OUTPUT FORMAT
The first line of the output should be the length of the longest palindrome found. The next line or lines should be the actual text of the palindrome (without any surrounding white space or punctuation but with all other characters) printed on a line (or more than one line if newlines are included in the palindromic text). If there are multiple palindromes of longest length, output the one that appears first.
SAMPLE OUTPUT (file calfflac.out)
11 Madam, I'm Adam
題目挺簡單，求字符串的最大回文子串，但是子串中可能有噪音(非字符符號、數字、空格、換行等)。要輸出迴文子串的長度(去掉非字母字符後的)以及迴文子串(將噪音還原後輸出)。
最長迴文子串，第一個念頭是Brute Force。。。但是麻煩，而且時間複雜度大概爲：N^3
後選擇DP。
直接上代碼
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
public class calfflac {
/**
* @param args
* @throws IOException
* calculate the longest palindrome substring length and output
* the original subString.
*/
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new FileReader("calfflac.in"));
FileWriter fout = new FileWriter("calfflac.out");
int n = br.read();
// String lines = br.readLine();
StringBuffer sb = new StringBuffer();
while (n != -1) {
sb.append((char) n);
n = br.read();
}
String sF = sb.toString();
System.out.println(sF.length());
String sR = sb.reverse().toString();
sb.reverse();
long start = System.currentTimeMillis();
int[] result = dp_lcss(sF, sR);
long end = System.currentTimeMillis();
System.out.println(end-start);
StringBuffer sb2 = new StringBuffer();
for(int i = 0;i<result[0]&&result[1]>0;) {
sb2.append(sR.charAt(result[1]-1));
if(isAlpha(sR.charAt(result[1]-1))) {
i++;
}
result[1]--;
}
fout.write(result[0]+"\n");
fout.write(sb2.toString() + "\n");
fout.flush();
fout.close();
br.close();
System.exit(0);
}
// modified dp longest common substring
public static int[] dp_lcss(String a, String b) {
long start = System.currentTimeMillis();
if (a == null || b == null) {
return new int[] {0,0};
}
int[] section = new int[2];
a = " " + a;
b = " " + b;
int[][] c = new int[a.length()][b.length()];
long end = System.currentTimeMillis();
for (int i = 0; i < a.length(); i++) {
c[i][0] = 0;
}
for (int j = 0; j < b.length(); j++) {
c[0][j] = 0;
}
int max = 0;
int interval_a = 1;
int interval_b = 1;
int cal_tmp = 0;
for (int i = 1; i < a.length();i++) {
if (!isAlpha(a.charAt(i))) {
interval_a++;
continue;
}
for (int j = 1; j < b.length();j++) {
if (!isAlpha(b.charAt(j))) {
interval_b++;
continue;
}
cal_tmp = Math.abs(a.charAt(i) - b.charAt(j));
if (cal_tmp==0||cal_tmp==32) {
c[i][j] = c[i - interval_a][j - interval_b] +1;
if (c[i][j] > max) {
max = c[i][j];
section[0] = max;
section[1] = j;
}
} else {
c[i][j] = 0;
}
interval_b = 1;
}
interval_a = 1;
interval_b = 1;
}
System.out.println(end-start);
return section;
}
private static boolean isAlpha(char c) {
// TODO Auto-generated method stub
if ((c <= 'z' && c >= 'a') || (c <= 'Z' && c >= 'A')) {
return true;
}
return false;
}
}中間參量interval_a和interval_b爲跳躍間隔，當出現非字母字符時，進行跳躍。
測試結果都沒問題，就是CPU時間超了。其中提供了一組測試數據，爲一個近4000字節的代碼片段，本機測試爲200多ms，但是提交之後就會達到4s，通常都再1.4s左右。看來只能使用後綴樹等其他方法進行計算，O(N^2)已經不能滿足它了。。。
又思維定勢了，剛參考了別人的解題思路，用的是我在不會DP時採用的笨方法：雙向比較的方法。等面試過後手寫下程序看看複雜度如何，還有就是證明下在下次循環時以上次比較的邊界爲起點的正確性。
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
public class calfflac {
/**
* @param args
* @throws IOException
* calculate the longest palindrome substring length and output
* the original subString.
*/
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
long start = System.currentTimeMillis();
BufferedReader br = new BufferedReader(new FileReader("calfflac.in"));
FileWriter fout = new FileWriter("calfflac.out");
int n = br.read();
// String lines = br.readLine();
StringBuffer sb = new StringBuffer();
while (n != -1) {
sb.append((char) n);
n = br.read();
}
// String sF = sb.toString();
System.out.println(sb.length());
// String sR = sb.reverse().toString();
// sb.reverse();
// int[] result = dp_lcss(sF, sR);
int result[] = bd_greedy(sb.toString());
StringBuffer sb2 = new StringBuffer();
/*for(int i = 0;i<result[0]&&result[1]>0;) {
sb2.append(sR.charAt(result[1]-1));
if(isAlpha(sR.charAt(result[1]-1))) {
i++;
}
result[1]--;
}*/
System.out.println(result[0]+" "+result[1]);
int i = result[0]/2;
int j = (result[0]+1)/2;
int tmp =result[1];
while(i>=0&&tmp>0){
if(isAlpha(sb.charAt(i)))
tmp--;
sb2.insert(0,sb.charAt(i));
i--;
}
if((result[0]&1)==0&&isAlpha(sb.charAt(result[0]/2))){
j++;
tmp =result[1]-1;
}else{
tmp =result[1];
}
while(j<sb.length()&&tmp>0){
if(isAlpha(sb.charAt(j)))
tmp--;
sb2.append(sb.charAt(j));
j++;
}
fout.write(result[1]*2-(1-(result[0]&1))+"\n");
fout.write(sb2.toString() + "\n");
fout.flush();
fout.close();
br.close();
long end = System.currentTimeMillis();
System.out.println(end-start);
System.exit(0);
}
private static int[] bd_greedy(String sF) {
// TODO Auto-generated method stub
int[] section = new int[2];
if (sF == null ) {
return section;
}
int counter = 0;
int i=0;
int j = 0;
for(int k =0;k<sF.length()*2-1&&(j<sF.length());k++){
i = k/2;
j=(k+1)/2;
while(i>=0&&j<sF.length()){
while(i>=0&&!isAlpha(sF.charAt(i)))
i--;
while(j<sF.length()&&!isAlpha(sF.charAt(j)))
j++;
if(i==-1||j==sF.length()||!isEqual(sF, i, j)){
break;
}
counter++;
i--;
j++;
}
if(section[1]<counter){
section[0] = k;
section[1] = counter;
}
counter = 0;
}
return section;
}
private static boolean isEqual(String sF, int i, int j) {
int cal_tmp = Math.abs(sF.charAt(i) - sF.charAt(j));
if(cal_tmp==0||cal_tmp==32)
return true;
return false;
}
// modified dp longest common substring
public static int[] dp_lcss(String a, String b) {
long start = System.currentTimeMillis();
int[] section = new int[2];
if (a == null || b == null) {
return section;
}
a = " " + a;
b = " " + b;
int[][] c = new int[a.length()][b.length()];
int interval_a = 1;
int interval_b = 1;
int cal_tmp = 0;
for (int i = 1; i < a.length();i++) {
if (!isAlpha(a.charAt(i))) {
interval_a++;
continue;
}
for (int j = 1; j < b.length();j++) {
if (!isAlpha(b.charAt(j))) {
interval_b++;
continue;
}
cal_tmp = Math.abs(a.charAt(i) - b.charAt(j));
if (cal_tmp==0||cal_tmp==32) {
c[i][j] = c[i - interval_a][j - interval_b] +1;
if (c[i][j] > section[0]) {
section[0] = c[i][j];
section[1] = j;
}
}
interval_b = 1;
}
interval_a = 1;
interval_b = 1;
}
long end = System.currentTimeMillis();
System.out.println(end-start);
return section;
}
private static boolean isAlpha(char c) {
// TODO Auto-generated method stub
if ((c <= 'z' && c >= 'a') || (c <= 'Z' && c >= 'A')) {
return true;
}
return false;
}
}使用i和j記錄每次進行迴文判斷到左右起點。方法輸出迴文字串的中心座標以及迴文長度。
時間複雜度：O(N^2)。最糟糕情況爲：所有字符都相同，即最大回文字串爲字符串本身。當已判斷的迴文子串長大於剩餘字符串長度到兩倍，可以停止搜索，以減少不必要操作。
