Edit Distance

难度：2

求编辑距离。。。

opt[i][j]表示word1的前i个字符与word2的前j个字符的editdistance
word1[i-1] == word2[j-1]  opt[i][j]=opt[i-1][j-1]
word1[i-1] != word2[j-1]  opt[i][j]=min(opt[i-1][j]+1,opt[i][j-1]+1,opt[i-1][j-1]+1)

Given two words word1 and word2, find the minimum number of steps required to convert word1 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

class Solution
{
public:
    int minDistance(string word1, string word2)
    {
        int m=word1.size();
        int n=word2.size();
        //opt[i][j]表示word1的前i个字符与word2的前j个字符的editdistance
        //word1[i-1] == word2[j-1]  opt[i][j]=opt[i-1][j-1]
        //word1[i-1] != word2[j-1]  opt[i][j]=min(opt[i-1][j]+1,opt[i][j-1]+1,opt[i-1][j-1]+1)
        int **opt=new int*[2];
        opt[0] = new int[n+1];
        opt[1] = new int[n+1];
        for(int i=0;i<=n;i++)
        {
            opt[0][i]=i;
        }
        for(int i=1,now=1;i<=m;i++,now^=1)
        {
            opt[now][0]=i;
            for(int j=1;j<=n;j++)
            {
                if(word1[i-1] == word2[j-1])    opt[now][j]=opt[now^1][j-1];
                else
                {
                    opt[now][j]=opt[now^1][j]+1;
                    opt[now][j]=min(opt[now][j],opt[now][j-1]+1);
                    opt[now][j]=min(opt[now][j],opt[now^1][j-1]+1);
                }
            }
        }
        int ans=opt[m&1?1:0][n];
        delete [] opt[0];
        delete [] opt[1];
        delete [] opt;
        return ans;
    }
};