思路
状态：dp[i][j]表示word1的前i个字符变成word2的前j个字符的最少操作数。
状态转移方程
word[i]==word[j]–>dp[i][j] = dp[i-1][j-1]word[i]!=word[j]需要考察，修改，插入，删除三种情况取最小值
f[i][j] = min(f[i-1][j-1],min(f[i][j-1],f[i-1][j]))+1
AC代码
C++
class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size();
vector<vector<int>> dp(m+1, vector<int> (n+1, 0));
for(int i = 0; i <= m; i++){
dp[i][0] = i;
}
for(int i = 0; i <= n; i++){
dp[0][i] = i;
}
for(i = 1; i <= m; i++){
for(j = 1; j <= n; j++){
if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = min(min(dp[i-1][j], dp[i-1][j-1]), dp[i][j-1]) + 1;
}
}
return dp[m][n];
}
};
Python
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m, n = len(word1), len(word2)
dp = [[0] * (n+1) for _ in range(m+1)]
for i in range(1, m+1):
dp[i][0] = i
for j in range(1, n+1):
dp[0][j] = j
for p in range(1, m+1):
for q in range(1, n+1):
if word1[p-1] == word2[q-1]:
dp[p][q] = dp[p-1][q-1]
else:
dp[p][q] = min(dp[p-1][q]+1, dp[p][q-1]+1, dp[p-1][q-1]+1)
return dp[-1][-1]