五、完善程序-2
(编辑距离)给定两个字符串,每次操作可以选择删除(Delete)、插入(Insert)、替换(Replace),一个字符,求将第一个字符串转换为第二个字符串所需要的最少操作次数。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | #include <iostream> #include <string> #include <vector> using namespace std; int min(int x, int y, int z) { return min(min(x, y), z); } int edit_dist_dp(string str1, string str2) { int m = str1.length(); int n = str2.length(); vector<vector<int>> dp(m + 1, vector<int>(n + 1)); for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i == 0) dp[i][j] = ①; else if (j == 0) dp[i][j] = ②; else if (③) dp[i][j] = ④; else dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j], ⑤); } } return dp[m][n]; } int main() { string str1, str2; cin >> str1 >> str2; cout << "Mininum number of operation:" << edit_dist_dp(str1, str2) << endl; return 0; } |
