交错字符串LCS

Question

嗨，我正试图解决交错字符串的问题。这是对这个问题的详细解释。https://practice.geeksforgeeks.org/problems/interleaved-strings/1我试着使用lcs，但它没有通过leetcode案例。这是我的代码：-(我从开始到结束服用lcs(

class Solution {
public boolean isInterLeave(String a, String b, String c) {

StringBuffer s=new StringBuffer();
StringBuffer s1=new StringBuffer();
StringBuffer s2=new StringBuffer();
StringBuffer s4=new StringBuffer();

int m=a.length();
int n=c.length();
int q=b.length();
if(n!=m+q){
return false;
}

LinkedHashSet<Integer> res2=  new LinkedHashSet<Integer>();
res2= lcs(a,c,m,n);
LinkedHashSet<Integer> res4=  new LinkedHashSet<Integer>();
res4= lcs(b,c,q,n);
for(int i=0;i<n;i++){
if(res2.contains(i)==false){
s.append(c.charAt(i));
}
}
for(int i=0;i<n;i++){
if(res4.contains(i)==false){
s1.append(c.charAt(i));
}
}
LinkedHashSet<Integer> res5=  new LinkedHashSet<Integer>();
res5= LCS(a,c,m,n);
for(int i=0;i<n;i++){
if(res5.contains(i)==false){
s2.append(c.charAt(i));
}
} LinkedHashSet<Integer> res6=  new LinkedHashSet<Integer>();
res6= LCS(b,c,q,n);
for(int i=0;i<n;i++){
if(res6.contains(i)==false){
s4.append(c.charAt(i));
}
}
String z=s.toString();
String u=s1.toString();
String v=s2.toString();
String w=s4.toString();

if( (b.equals(z)==true || a.equals(u)==true) || ( b.equals(v)==true || a.equals(w)==true)){
return true;
}

else{
return false;
}  
}
public static LinkedHashSet<Integer> lcs(String X, String Y, int m, int n)
{
int[][] L = new int[m+1][n+1];
// Following steps build L[m+1][n+1] in bottom up fashion. Note
// that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
}
}
// Following code is used to print LCS

// Create a character array to store the lcs string
LinkedHashSet<Integer> linkedset = 
new LinkedHashSet<Integer>();  
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i=1;
int j=1;
while (i <= m && j <= n)
{
// If current character in X[] and Y are same, then
// current character is part of LCS
if (X.charAt(i-1) == Y.charAt(j-1))
{
// Put current character in result

linkedset.add(j-1);
// reduce values of i, j and index
i++;
j++;

}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[i-1][j] > L[i][j-1])
i++;
else
j++;
}

return linkedset;

}
public static LinkedHashSet<Integer> LCS(String X, String Y, int m, int n)
{
int[][] L = new int[m+1][n+1];
// Following steps build L[m+1][n+1] in bottom up fashion. Note
// that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
}
}
// Following code is used to print LCS

// Create a character array to store the lcs string
LinkedHashSet<Integer> linkedset = 
new LinkedHashSet<Integer>();  
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i = m;
int j = n;
while (i > 0 && j > 0)
{
// If current character in X[] and Y are same, then
// current character is part of LCS
if (X.charAt(i-1) == Y.charAt(j-1))
{
// Put current character in result
linkedset.add(j-1); 

// reduce values of i, j and index
i--; 
j--; 

}

// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[i-1][j] > L[i][j-1])
i--;
else
j--;
}
return linkedset;

}
}

有人能提出LCS解决这个问题的方法吗？。我的代码没有通过以下测试用例"cacabcbaccbbcbb"-字符串A"acaaccaacbbbabbacc"-字符串B"accacaabcbacacacbbbbcbabbbacc"-字符串C

Answer 1

这将是LCS+DP方法。试试看：

class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
int m = s1.length(), n = s2.length();
if (n + m != s3.length()) return false;
if (s3.length() == 0) return true;

boolean[][] dp = new boolean[m+1][n+1];
dp[0][0] = true;
for (int i = 0; i <= m; i++) {
if (s1.substring(0, i).equals(s3.substring(0, i))) 
dp[i][0] = true;
else 
dp[i][0] = false;
}
for (int j = 0; j <= n; j++) {
if (s2.substring(0, j).equals(s3.substring(0, j))) 
dp[0][j] = true;
else 
dp[0][j] = false;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
dp[i][j] = (dp[i-1][j] && s1.charAt(i-1) == s3.charAt(i+j-1)) 
|| (dp[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1));
}
}
return dp[m][n];
}
}