Edit Distance（两字符串经过最少操作匹配）

72. Edit Distance

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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

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显然采用动态规划做：

用dp[n+1][m+1]表示需要执行的最少变换次数，则递推方程为：

• dp[i+1][j+1] = dp[i][j],                  当word1[i] == word2[j]
• 
  
• dp[i + 1][j + 1] = min(min(dp[i][j] + 1, dp[i][j + 1] + 1), dp[i + 1][j] + 1);  当word1[i] != word2[j]
  

class Solution
{
public:
    int minDistance(string word1, string word2)
    {
        int len1 = word1.size(), len2 = word2.size();
        
        vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1));
        
        // 左边沿初始化
        for (int i = 0; i <= len1; ++i)
            dp[i][0] = i;
        
        // 右边沿初始化
        for (int i = 0; i <= len2; ++i)
            dp[0][i] = i;
        
        for (int i = 0; i < len1; ++i)
        {
            for (int j = 0; j < len2; ++j)
            {
                if (word1[i] == word2[j])
                    dp[i + 1][j + 1] = dp[i][j];
                else
                    dp[i + 1][j + 1] = min(min(dp[i][j] + 1, dp[i][j + 1] + 1), dp[i + 1][j] + 1);
            }
        }
        return dp[len1][len2];
    }
};