代码随想录Day55

class Solution {
public:
// 动态规划五部曲：dp定义、初始化、递推公式、递归顺序、推导
    int minDistance1(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,min( dp[i][j - 1] + 1, dp[i - 1][j - 1] + 2));
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
    // 法二、求最长公共子序列，然后把除了该子序列以外的元素都删除
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
        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] + 1;
                } else {
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return (word1.size() + word2.size() - 2 * dp[word1.size()][word2.size()]);
    }
};

class Solution {
public:
    int minDistance(string word1, string word2) {
        int w1 = word1.size();
        int w2 = word2.size();
        vector<vector<int>> dp(w1 + 1, vector<int>(w2 + 1));
        for (int i = 0; i <= w1; ++i) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= w2; ++j) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= w1; ++i) {
            for (int j = 1; j <= w2; ++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],min( dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                }
            }
        }
        return dp[w1][w2];
    }
};