583. 两个字符串的删除操作
可以求出最大子序列然后用字符串长度去减,也可以用删除的思路,如下:
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++){dp[i][0]=i;}for(int i=1;i<=word2.size();i++){dp[0][i]=i;}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.back().back();}
};
相等就和两个对比之前的串不变,不相等可以模拟一个串删除一个字符的操作
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=1;i<=word1.size();i++){dp[i][0]=i;}for(int i=1;i<=word2.size();i++){dp[0][i]=i;}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][j-1]+1,dp[i-1][j]+1));}}}return dp.back().back();}
};
min里分别是,替换,插入,删除