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动态规划part16
LeetCode583两个字符串的删除操作
class Solution { public: int minDistance(string word1, string word2) { vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1)); for (int i = 0; i <= word1.size(); i++) dp[i][0] = i; for (int j = 0; j <= word2.size(); j++) dp[0][j] = j; for (int i = 1; i <= word1.size(); i++) { for (int j = 1; j <= word2.size(); j++) { if (word1[i - 1] == word2[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); } } } return dp[word1.size()][word2.size()]; } };
LeetCode72编辑距离
class Solution { public: int minDistance(string word1, string word2) { vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0)); for (int i = 0; i <= word1.size(); i++) dp[i][0] = i; for (int j = 0; j <= word2.size(); j++) dp[0][j] = j; for (int i = 1; i <= word1.size(); i++) { for (int j = 1; j <= word2.size(); j++) { if (word1[i - 1] == word2[j - 1]) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1; } } } return dp[word1.size()][word2.size()]; } };
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