Edit Distance

来源：互联网发布：java io importnew 编辑：程序博客网时间：2024/04/27 02:47

1.对于f[i][j]可以由三种状态而来，即

f[i-1][j-1]：修改word1[i]=word2[j]

f[i-1][j]：删除word2[j]或者插入word1[i]

f[i][j-1]：删除word1[i]或者插入word2[j]

2.状态方程如下

$\\if ~~word1[i]==word2[j]\\ ~~~~~f[i][j]=f[i-1][j-1]\\ else\\ ~~~~~f[i][j]=min(f[i-1][j-1],f[i][j-1],f[i-1][j])+1\\$

3.例如，对于字符串abcd 和acbd，填表如下

$\begin{matrix} \cdot&\cdot &a &b &c &d \\ \cdot&0 &1 &2 &3 &4 \\ a&1 &0 &1 &2 &3 \\ c&2 &1 &1 &1 &2 \\ b&3 &2 &1 &2 &2 \\ d&4 &3 &2 &2 &2 \end{matrix}$

class Solution {public:    int minDistance(string word1, string word2) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int len1=word1.length(),len2=word2.length(),i,j;        if(len1==0)return len2;        if(len2==0)return len1;        vector<int> row(len2+1,0);        vector<vector<int>> f(len1+1,row);        for(i=0;i<=len1;++i){            f[i][0]=i;        }        for(i=0;i<=len2;++i){            f[0][i]=i;        }        for(i=1;i<=len1;++i){            for(j=1;j<=len2;++j){                if(word1[i-1]==word2[j-1]){                    f[i][j]=f[i-1][j-1];                }else{                    //          replace         delete or insert                    f[i][j]=min(f[i-1][j-1],min(f[i-1][j],f[i][j-1]))+1;                }            }        }        return f[len1][len2];    }};