文章链接:代码随想录
题目链接:583. 两个字符串的删除操作
思路:直接记录需要改(增或删)几个,也就是求不公共的子序列
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] + 2, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1));
}
}
return dp[word1.size()][word2.size()];
}
};
也可以记录最长公共子序列,再减
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
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][j - 1], dp[i - 1][j]);
}
}
return word1.size() + word2.size() - dp[word1.size()][word2.size()] * 2;
}
};
思路:和上一题相比,差别在于多了替换,因此dp[i - 1][j - 1] 只需要多加一步即可变为dp[i][j]。
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 = 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 - 1][j - 1] + 1, min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
}
}
return dp[word1.size()][word2.size()];
}
};
第五十六天打卡,今天给周老师写了个冰层项目进展,耽误了一些学习进度,加油!!!