leetcode72

 class Solution {
 private:
     int d[][];

 public:
     int minDistance(string word1, string word2)
     {
         int len1 = word1.length();
         int len2 = word2.length();

         for(int i=;i<=len1;i++)
             d[i][]= i;

         for(int j=;j<=len2;j++)
             d[][j]=j;

         for(int i=;i <=len1;i++)
         {
             for(int j=;j<=len2;j++)
             {
                 int diff;
                 if(word1[i-] == word2[j-])
                     diff =  ;
                 else
                     diff =  ;
                 int temp = min(d[i-][j] + , d[i][j-] + );
                 d[i][j] = min(temp, d[i-][j-] + diff);
             }
         }
         return d[len1][len2];
     }
 };

这是《趣学算法》一书中4-4节提供的代码，在leetcode上运行，速度比较慢，392ms（5.16%），12.9mb（5.22%）。

补充一个python的实现：

 class Solution:
     def minDistance(self, word1: str, word2: str) -> int:
         m = len(word1)
         n = len(word2)
         dp = [[ for _ in range(n+)]for _ in range(m+)]
         for j in range(n+):
             dp[][j] = j
         for i in range(m+):
             dp[i][] = i
         for i in range(,m+):
             for j in range(,n+):
                 diff =  if word1[i-] != word2[j-] else
                 minval = min(dp[i-][j]+,dp[i][j-]+)
                 minval = min(minval,dp[i-][j-]+diff)
                 dp[i][j] = minval
         return dp[m][n]

240ms，16.4mb

下面提供leetcode中一个12ms，8.5mb的解决方案：

 class Solution {
 public:
     int minDistance(string word1, string word2) {
         int m = word1.length();
         int n = word2.length();
         vector<int> dp(m+, );
         int i, j;
         int temp, min;
         for(i=;i<=m;i++)
             dp[i] = i;
         for(j=;j<=n;j++)
         {
             temp = dp[];
             dp[]++;
             for(i=;i<=m;i++)
             {
                 min = temp + (word1[i-]!=word2[j-]);
                 min = (dp[i-]+)<min?(dp[i-]+):min;
                 min = (dp[i]+)<min?(dp[i]+):min;
                 temp = dp[i];
                 dp[i] = min;
             }
         }
         return dp[m];
     }
 };

将第二种写法，转化为python语法：

 class Solution:
     def minDistance(self, word1: str, word2: str) -> int:
         m = len(word1)
         n = len(word2)
         dp = list(range(m+))
         for j in range(,n+):
             temp = dp[]
             dp[] +=
             for i in range(,m+):
                 diff =  if word1[i-] != word2[j-] else
                 minval = temp + diff
                 minval = min(minval,dp[i-]+)
                 minval = min(minval,dp[i]+)
                 temp = dp[i]
                 dp[i] = minval
         return dp[m]

204ms，12.9mb