代码随想录算法训练营第五十八天| 583. 两个字符串的删除操作 、72. 编辑距离
- 583. 两个字符串的删除操作
- 72. 编辑距离
哎,最近的题有点难啊~~~~~
583. 两个字符串的删除操作
题目链接:583. 两个字符串的删除操作
文章链接
状态:不太会,直接看的解
代码
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
题目链接:72. 编辑距离
文章链接
状态:不会做嘤嘤嘤
代码
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()];}
};