代码随想录算法训练营第40天 | 第九章 动态规划12
文章目录
- 今日记录
- 115.不同的子序列
- 583. 两个字符串的删除操作
- 72. 编辑距离
- 总结
今日记录
115.不同的子序列
Leetcode链接
class Solution {
public:int numDistinct(string s, string t) {vector<vector<uint64_t>> dp(s.size() + 1, vector<uint64_t>(t.size() + 1, 0));for (int i = 0; i <= s.size(); i++) {dp[i][0] = 1;}for (int j = 1; j <= t.size(); j++) {dp[0][j] = 0;}for (int i = 1; i <= s.size(); i++) {for (int j = 1; j <= t.size(); j++) {if (s[i - 1] == t[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];} else {dp[i][j] = dp[i - 1][j];}}}return dp[s.size()][t.size()];}
};
583. 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, 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() - dp[word1.size()][word2.size()] * 2;}
};
72. 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 = 1; 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][j - 1], dp[i - 1][j], dp[i - 1][j - 1]}) + 1;}}}return dp[word1.size()][word2.size()];}
};