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

 思路：题意求字符串的最小编辑距离。设状态为f[i][j],表示A[0][i] B[0][j]之间的最小编辑距离。

forexamplr:str1c str2d

1、c==d f[i][j]=f[i-1][j-1] 2、c!=d 　 (1)如果将c换成d 则f[i][j]=f[i-1][j-1]+1 (2)如果在c后面添加一个d f[i][j]=f[i][j-1]+1 (3)如果将c删除 f[i][j]=f[i-1][j-1]+1

   简单的状态方程为

dp[i, j] = min { dp[i - 1, j] + 1,  dp[i, j - 1] + 1, dp[i - 1, j - 1] + (s[i] == t[j] ? 0 : 1) }

1 class Solution { 2 public: 3     int minDistance(string word1, string word2) { 4          5         const int n=word1.size(); 6         const int m=word2.size(); 7          8         vector
     
      
       > f(n+1,vector
       
        (m+1,0)); 9         10         for(int i=0;i<=n;i++)11             f[i][0]=i;12             13         for(int j=0;j<=m;j++)14             f[0][j]=j;15             16         for(int i=1;i<=n;i++)17             for(int j=1;j<=m;j++)18                 if(word1[i-1]==word2[j-1])19                     f[i][j]=f[i-1][j-1];20                 else{21                     f[i][j]=min(f[i-1][j-1],min(f[i-1][j],f[i][j-1]))+1;22                     23                 }24     25         return f[n][m];26     }27 };