class Solution { public: /** * min edit cost * @param str1 string字符串 the string * @param str2 string字符串 the string * @param ic int整型 insert cost * @param dc int整型 delete cost * @param rc int整型 replace cost * @return int整型 */ int minEditCost(string str1, string str2, int ic, int dc, int rc) { // write code here int len1 = str1.size(); int len2 = str2.size(); vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0)); for (int i = 1; i <= len1; i++) dp[i][0] = i * dc; for (int i = 1; i <= len2; i++) dp[0][i] = i * ic; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1[i - 1] == str2[j - 1]) dp[i][j] = dp[i - 1][j - 1]; //r1[i] = str2[j] else { dp[i][j] = min({dp[i - 1][j] + dc, dp[i - 1][j - 1] + rc, dp[i][j - 1] + ic}); //dp[i][j] 取三种措施的最小的代价 } } } return dp[len1][len2]; } };