【leetcode】Edit Distance

Edit Distance

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

采用动态规划求解

1. d[0, j] = j;

2. d[i, 0] = i;

3. d[i, j] = d[i-1, j - 1] if A[i] == B[j]

4. d[i, j] = min(d[i-1, j - 1], d[i, j - 1], d[i-1, j]) + 1  if A[i] != B[j]

 

 class Solution {
 public:
     int minDistance(string word1, string word2) {

         int n1=word1.length();
         int n2=word2.length();

         if(n1==) return n2;
         if(n2==) return n1;

         //采用二维数组效率更高
         //vector<vector<int> > dp(n1+1,vector<int>(n2+1));

         int **dp=new int*[n1+];
         for(int i=;i<n1+;i++)
         {
             dp[i]=new int[n2+];
         }

         dp[][]=;
         for(int i=;i<=n1;i++)
         {
             dp[i][]=i;
         }
         for(int j=;j<=n2;j++)
         {
             dp[][j]=j;
         }

         for(int i=;i<=n1;i++)
         {
             for(int j=;j<=n2;j++)
             {

                 if(word1[i-]==word2[j-])
                 {
                     dp[i][j]=dp[i-][j-];
                 }
                 else
                 {
                     dp[i][j]=min(dp[i-][j],min(dp[i][j-],dp[i-][j-]))+;
                 }
             }
         }

         int result=dp[n1][n2];
         for(int i=;i<n1+;i++)
         {
             delete[] dp[i];
         }

         return result;
     }
 };