Edit Distance

题目：

Given two words word1 and word2, find the minimum number of steps required to convertword1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

分析：可以利用以下步骤

1Set 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].

比较"GUMBO"和"GAMBOL"的相似程度，计算两者的edit distance

 代码如下：

    int min(int a,int b,int c)
    {
        int tmp1=a<b?a:b;
        return tmp1<c?tmp1:c;
        
    }
    int minDistance(string word1, string word2)
    {
        int n=word1.length();
        int m=word2.length();
        if(n==0)return m;
        if(m==0)return n;
        int **matrix=new int *[m+1];
        for(int i=0;i<=m;i++)
        {
            matrix[i]=new int[n+1];
        }
        for(int i=0;i<=m;i++)
        {
            matrix[i][0]=i;
        }
        for(int i=1;i<=n;i++)
        {
            matrix[0][i]=i;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                //int tmp=min(matrix[j-1][i],matrix[j][i-1],matrix[j-1][i-1]);
                if(word1[i-1]==word2[j-1])
                {
                    matrix[j][i]=min(matrix[j-1][i]+1,matrix[j][i-1]+1,matrix[j-1][i-1]);
                }
                else
                {
                    matrix[j][i]=min(matrix[j-1][i]+1,matrix[j][i-1]+1,matrix[j-1][i-1]+1);
                }              
            }
        }
        int result=matrix[m][n];
        for(int i=0;i<=m;i++)
        {
            delete []matrix[i];
        }
        delete []matrix;
        return result;
    }