leetCode 72. Edit Distance-优快云博客

本文链接：https://blog.youkuaiyun.com/liujunzxcv/article/details/78049274

本文介绍了一种使用动态规划方法来计算两个字符串之间的最小编辑距离。通过插入、删除或替换字符的操作，实现了从一个字符串转换为另一个字符串所需的最少步骤计算。

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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

计算word1变换到Word2的距离，典型的动态规划，本来想直接递归做，但是超时。

 public static int minDistance(String word1, String word2) {
         if(word1==null && word2==null )
             return 0;

         if(word1 == null || word1.length()<1)return word2.length();
         if(word2 == null || word2.length()<1)return word1.length();
         if(word1.length()==1&&word2.length()==1){
             if(word1.equals(word2))
                 return 0;
             else return 1;
         }

         if(word1.equals(word2))return 0;
         int row = word1.length();
         int col = word2.length();
         int [][]A = new int[row+1][col+1];
         for(int i=0;i<=row;i++){
             A[i][0]=i;
         }
         for(int i=0;i<=col;i++){
             A[0][i]=i;
         }
         for(int i=1;i<=row;i++){
             for(int j=1;j<=col;j++){
                 if(word1.charAt(i-1)==word2.charAt(j-1))
                 //当word1[i]==word2[j]，直接就等于A[i-1][j-1]
                     A[i][j]=min(A[i-1][j]+1,A[i][j-1]+1,A[i-1][j-1]);
                 else
                     A[i][j]=min(A[i-1][j]+1,A[i][j-1]+1,A[i-1][j-1]+1);
             } 
         }
         return A[row][col];
     }


     public static int min(int i,int j,int k){
         int z = (i<j?i:j);
         return k<z?k:z;
     }