题解 | #计算字符串的编辑距离#
计算字符串的编辑距离
https://www.nowcoder.com/practice/3959837097c7413a961a135d7104c314
动态规划问题。dp数组dp[i][j]表示word1中前i个字符，变换到word2中前j个字符所使用的最少操作数。要达到dp[i][j]有三种操作，增加、删除、和修改。三者取最小。
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
string s1, s2;
getline(cin, s1, '\n');
getline(cin, s2, '\n');
vector<vector<int>> dp(s1.size() + 1, vector<int>(s2.size() + 1, 0));
// 初始化
for (int i = 0; i <= s1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= s2.size(); j++) dp[0][j] = j;
for (int i = 1; i <= s1.size(); i++) {
for (int j = 1; j <= s2.size(); j++) {
if (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i-1][j-1];
else {
dp[i][j] = min({dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + 1});
}
}
}
cout << dp[s1.size()][s2.size()] << endl;
return 0;
}
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