f[i,j] A 1-i 变成 B 1-j 的最短编辑距离
增,删,改
改分两种情况,是否相同
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
if (A[i] == B[j]) {
f[i][j] = min(f[i][j], f[i - 1][j - 1]);
} else {
f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
}
#include <iostream>
using namespace std;
const int N = 1010;
char A[N], B[N];
int f[N][N];
int main() {
int n, m;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> A[i];
}
cin >> m;
for (int i = 1; i <= m; ++i) {
cin >> B[i];
}
for (int i = 0; i <= n; ++i) {
f[i][0] = i;
}
for (int i = 0; i <= m; ++i) {
f[0][i] = i;
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
if (A[i] == B[j]) {
f[i][j] = min(f[i][j], f[i - 1][j - 1]);
} else {
f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
}
}
}
cout << f[n][m] << endl;
}