字符串相似度算法（ Levenshtein Distance算法）題目： 一個字符串可以通過增加一個字符，刪除一個字符，替換一個字符得到另外一個字符串，假設(shè)，我們把從字符串A轉(zhuǎn)換成字符串B，前面3種操作所執(zhí)行的最少次數(shù)稱為AB相似度
C++實現(xiàn)如下  #include <iostream>#include <vector>#include <string>using namespace std;//算法int ldistance(const string source,const string target)   { //step 1 int n=source.length(); int m=target.length(); if (m==0) return n; if (n==0) return m; //Construct a matrix typedef vector< vector<int> > Tmatrix; Tmatrix matrix(n+1); for(int i=0; i<=n; i++) matrix[i].resize(m+1); //step 2 Initialize for(int i=1;i<=n;i++) matrix[i][0]=i; for(int i=1;i<=m;i++) matrix[0][i]=i; //step 3 for(int i=1;i<=n;i++) { const char si=source[i-1]; //step 4 for(int j=1;j<=m;j++) { const char dj=target[j-1]; //step 5 int cost; if(si==dj){ cost=0; } else{ cost=1; } //step 6 const int above=matrix[i-1][j]+1; const int left=matrix[i][j-1]+1; const int diag=matrix[i-1][j-1]+cost; matrix[i][j]=min(above,min(left,diag)); } }//step7 return matrix[n][m];}int main(){ string s; string d; cout<<"source="; cin>>s; cout<<"diag="; cin>>d; int dist=ldistance(s,d); cout<<"dist="<<dist<<endl;}#include <iostream>#include <vector>#include <string>using namespace std;//算法int ldistance(const string source,const string target){ //step 1 int n=source.length(); int m=target.length(); if (m==0) return n; if (n==0) return m; //Construct a matrix typedef vector< vector<int> > Tmatrix; Tmatrix matrix(n+1); for(int i=0; i<=n; i++) matrix[i].resize(m+1); //step 2 Initialize for(int i=1;i<=n;i++) matrix[i][0]=i; for(int i=1;i<=m;i++) matrix[0][i]=i; //step 3 for(int i=1;i<=n;i++) { const char si=source[i-1]; //step 4 for(int j=1;j<=m;j++) { const char dj=target[j-1]; //step 5 int cost; if(si==dj){ cost=0; } else{ cost=1; } //step 6 const int above=matrix[i-1][j]+1; const int left=matrix[i][j-1]+1; const int diag=matrix[i-1][j-1]+cost; matrix[i][j]=min(above,min(left,diag)); } }//step7 return matrix[n][m];}int main(){ string s; string d; cout<<"source="; cin>>s; cout<<"diag="; cin>>d; int dist=ldistance(s,d); cout<<"dist="<<dist<<endl;}java 字符串編輯距離算法實現(xiàn)： public static int getLevenshteinDistance (String s, String t) {
if (s == null || t == null) {
throw new IllegalArgumentException("Strings must not be null");
}
/*
The difference between this impl. and the previous is that, rather
than creating and retaining a matrix of size s.length()+1 by t.length()+1,
we maintain two single-dimensional arrays of length s.length()+1. The first, d,
is the 'current working' distance array that maintains the newest distance cost
counts as we iterate through the characters of String s. Each time we increment
the index of String t we are comparing, d is copied to p, the second int[]. Doing so
allows us to retain the previous cost counts as required by the algorithm (taking
the minimum of the cost count to the left, up one, and diagonally up and to the left
of the current cost count being calculated). (Note that the arrays aren't really
copied anymore, just switched...this is clearly much better than cloning an array
or doing a System.arraycopy() each time through the outer loop.)
Effectively, the difference between the two implementations is this one does not
cause an out of memory condition when calculating the LD over two very large strings.
*/
int n = s.length(); // length of s
int m = t.length(); // length of t
if (n == 0) {
return m;
} else if (m == 0) {
return n;
}
int p[] = new int[n+1]; //'previous' cost array, horizontally
int d[] = new int[n+1]; // cost array, horizontally
int _d[]; //placeholder to assist in swapping p and d
// indexes into strings s and t
int i; // iterates through s
int j; // iterates through t
char t_j; // jth character of t
int cost; // cost
for (i = 0; i<=n; i++) {
p[i] = i;
}
for (j = 1; j<=m; j++) {
t_j = t.charAt(j-1);
d[0] = j;
for (i=1; i<=n; i++) {
cost = s.charAt(i-1)==t_j ? 0 : 1;
// minimum of cell to the left+1, to the top+1, diagonally left and up +cost
d[i] = Math.min(Math.min(d[i-1]+1, p[i]+1), p[i-1]+cost);
}
// copy current distance counts to 'previous row' distance counts
_d = p;
p = d;
d = _d;
}
// our last action in the above loop was to switch d and p, so p now
// actually has the most recent cost counts
return p[n];
}
字符串相似度=1-（編輯距離/（MAX（字符串1長度，字符串2的長度））
oracle 11提供了計算字符串編輯距離和相似度的函數(shù)： 參見http:///reference/utl_match.html
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