计算字符串相似度（）

来源：互联网发布：手机吹笛子软件编辑：程序博客网时间：2024/06/04 01:03

原文地址点击打开链接

题目：一个字符串可以通过增加一个字符，删除一个字符，替换一个字符得到另外一个字符串，假设，我们把从字符串A转换成字符串B，前面3种操作所执行的最少次数称为AB相似度
如  abc adc  度为 1
      ababababa babababab 度为 2
      abcd acdb 度为2

字符串相似度算法可以使用 Levenshtein Distance算法(中文翻译：编辑距离算法) 这算法是由俄国科学家Levenshtein提出的。其步骤

StepDescription1Set n to be the length of s.
Set m to be the length of t.
If n = 0, return m and exit.
If m = 0, return n and exit.
Construct a matrix containing 0..m rows and 0..n columns.2Initialize the first row to 0..n.
Initialize the first column to 0..m.3Examine each character of s (i from 1 to n).4Examine each character of t (j from 1 to m).5If s[i] equals t[j], the cost is 0.
If s[i] doesn't equal t[j], the cost is 1.6Set cell d[i,j] of the matrix equal to the minimum of:
a. The cell immediately above plus 1: d[i-1,j] + 1.
b. The cell immediately to the left plus 1: d[i,j-1] + 1.
c. The cell diagonally above and to the left plus the cost: d[i-1,j-1] + cost.7After the iteration steps (3, 4, 5, 6) are complete, the distance is found in cell d[n,m].

C++实现如下：

C/C++ code复制代码

#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;}