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81 lines (73 loc) · 1.94 KB
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package code_07_dp;
import org.junit.Test;
/**
* 求解LCS问题的具体解
*
* Sample Input:
abcfbc abfcab
programming contest
abcd mnp
* Sample Output:
"abcb"
2
""
*/
public class LCSSolution {
StringBuilder res;
//记录最长公共子序列的走向 'c'表示斜方向 ‘l’表示左方向 ‘u’表示上方向
char[][] x;
public int LCSLength(String s,String t,int m,int n){
res=new StringBuilder();
if(m==0 || n==0){
return 0;
}
int[][] lcs=new int[m+1][n+1];
x=new char[m+1][n+1];
for(int i=0;i<m+1;i++){
for(int j=0;j<n+1;j++){
lcs[i][j]=0;
}
}
for(int i=1;i<m+1;i++){
for(int j=1;j<n+1;j++){
//注意：这里的i是从1开始的 i=m时，实际上取到的是s的第m个元素
if(s.charAt(i-1)==t.charAt(j-1)){
lcs[i][j]=1+lcs[i-1][j-1];
x[i][j]='c';
}else if(lcs[i-1][j]>=lcs[i][j-1]){
lcs[i][j]=lcs[i-1][j];
x[i][j]='u';
}else if(lcs[i-1][j]<lcs[i][j-1]){
lcs[i][j]=lcs[i][j-1];
x[i][j]='l';
}
}
}
return lcs[m][n];
}
public void printLCS(String s,int m,int n){
if(m==0 || n==0){
return;
}
if(x[m][n]=='c'){
printLCS(s,m-1,n-1);
res.append(s.charAt(m-1));
}else if(x[m][n]=='u'){
printLCS(s,m-1,n);
}else if(x[m][n]=='l'){
printLCS(s,m,n-1);
}
}
@Test
public void test(){
String s="abcfbc";
String t="abfcab";
int m=s.length();
int n=t.length();
//String s="a";
//String t="ab";
LCSLength(s,t,m,n);
printLCS(s,m,n);
System.out.println(res.toString());
}
}