最长公共子序列
Description
给定两个字符串,返回两个字符串的最长公共子序列(不是最长公共子字符串),可能是多个。
Input
输入第一行为用例个数, 每个测试用例输入为两行,一行一个字符串
Output
如果没有公共子序列,不输出,如果有多个则分为多行,按字典序排序。
Sample Input 1
1
1A2BD3G4H56JK
23EFG4I5J6K7
Sample Output 1
23G456K
23G45JK
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class Q1_1 {
private static int T;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
T = in.nextInt();
in.nextLine();
while (T-- != 0) {
String a = in.nextLine();
String b = in.nextLine();
Set<String> lcsSet = LCS(a, b);
ArrayList<String> list = new ArrayList<>(lcsSet);
if (list.size() == 1 && list.get(0).equals("")){
continue;
}
list.sort(String::compareTo);
for (int i = 0; i < list.size(); i++) {
print(list.get(i), list.size()-1-i);
}
}
}
private static void print(String s, int n){
if (T == 0 && n == 0){
System.out.println(s);
}else{
System.out.println(s );
}
}
public static Set<String> LCS(String a, String b) {
int n = a.length(), m = b.length();
//c[i][j]表示a长度为i和b长度为j时的最长公共子序列长度
int[][] c = new int[n+1][m+1];
//d[i][j]表示方向
char[][] d = new char[n+1][m+1];
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
if(a.charAt(i-1) == b.charAt(j-1)) {
c[i][j] = c[i-1][j-1] + 1;
//左上
d[i][j] = '↖';
} else if(c[i-1][j] > c[i][j-1]){
c[i][j] = c[i-1][j];
//上
d[i][j] = '↑';
} else if(c[i-1][j] < c[i][j-1]){
c[i][j] = c[i][j-1];
//左
d[i][j] = '←';
} else {
c[i][j] = c[i][j-1];
//可向左可向右
d[i][j] = '┘';
}
// System.out.print(j == m ? c[i][j] + "\n":c[i][j] + " ");
}
}
// for(int i = 0; i <= n; ++i){
// for(int j = 0; j <= m; ++j) {
// System.out.print(j == m ? d[i][j] + "\n":d[i][j] + " ");
// }
// }
String lcs = "";
Set<String> lcsSet = new HashSet<>();
backTrace(d,a,lcs,n,m,c[n][m],lcsSet);
return lcsSet;
}
public static void backTrace(char[][] d, String a, String lcs, int i , int j, int maxSublen, Set<String> lcsSet){
if(i == 0 || j == 0) {
StringBuilder sb = new StringBuilder(lcs);
lcs = sb.reverse().toString();
//可能有些提早出来了,一定要判断长度是最长的,但是这样还是会有重复的字符串,所以还要做去重处理
if (lcs.length() == maxSublen) {
lcsSet.add(lcs);
}
return;
}
switch (d[i][j]){
case '↖':
lcs += a.charAt(i-1);
backTrace(d,a,lcs,i-1,j-1,maxSublen,lcsSet);
break;
case '↑':
backTrace(d,a,lcs,i-1,j,maxSublen,lcsSet);
break;
case '←':
backTrace(d,a,lcs,i,j-1,maxSublen,lcsSet);
break;
case '┘':
backTrace(d,a,lcs,i-1,j,maxSublen,lcsSet);
backTrace(d,a,lcs,i,j-1,maxSublen,lcsSet);
break;
}
}
}