代码随想录算法训练营day56

583. 两个字符串的删除操作

五部曲：

dp数组下标及含义：

• dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，想要达到相等，所需要删除元素的最少次数。

dp数组初始化：

• dp[i][0] = i；dp[0][j]=j。

递推公式：

• if (word1[i - 1] = word2[j - 1])，dp[i][j] = dp[i - 1][j - 1];
• min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});

遍历方向： 从前到后遍历

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()];
    }
};

72. 编辑距离

五部曲：

dp数组下标及含义：

• dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，最近编辑距离为dp[i][j]。

dp数组初始化：

• dp[i][0] = i；dp[0][j]=j。

递推公式：

• 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;

遍历方向： 从前到后遍历

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()];
    }
};