LeetCode 72 Edit Distance DP

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

Solution

设 \(dp[i][j]\) 表示将 word1的 \([0,...,i)\) 转变为 word2 的 \([0,...,j)\) 的最小操作数。对于初始化，即考虑某一个串为空串，操作数自然为另一个非空串的长度：

\[dp[i][0] = i;dp[0][j] = j \]

现考虑转移：假设现在已经知道了 \(dp[i][j]\) 之前的所有状态最小操作数，如果 \(s1[i]==s2[j]\)，则直接：

\[dp[i][j] = dp[i-1][j-1] \]

否则，则是考虑三个操作所带来的花费：

• 最简单的就是 replace: \(dp[i][j] = dp[i-1][j-1]+1\)
• 如果 \(s1[0,...,i-1) =s2[0,...,j)\)，则直接 delete \(s[i-1]\):

\[dp[i][j] = dp[i-1][j]+1 \]

• 如果 \(s1[0,...,i)+s2[j-1]=s2[0,...,j)\)，则进行 insert：

\[dp[i][j] = dp[i][j-1]+1 \]

点击查看代码

class Solution {
private:
    int dp[505][505];
public:
    int minDistance(string word1, string word2) {
        int n = word1.length(), m = word2.length();
        // dp[i][j]: word1[0,...,i) word2[0,...,j)
        for(int i=0;i<=n;i++)dp[i][0] = i;
        for(int i=0;i<=m;i++)dp[0][i] = i;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(word1[i-1]==word2[j-1])dp[i][j] = dp[i-1][j-1];
                else{
                    // word1[i-1]!=word2[j-1]
                    // dp[i-1][j-1] is known now
                    // Replace: dp[i][j] = dp[i-1][j-1]+1;
                    // Delete: dp[i][j] = dp[i-1][j]+1;
                    // Insert: dp[i][j] = dp[i][j-1]+1;
                    dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;
                }
            }
        }
        return dp[n][m];
    }
};