Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
class Solution { public: int minDistance(string word1, string word2) { //dp[i][j] 表示 ord1[0..i-1] -> word2[0..j-1] 最少的次数 int n1=word1.size(); int n2=word2.size(); vector<vector<int> > dp(n1+1,vector<int>(n2+1,0)); // 表示 [0..i]->[0..j] 的最少编辑次数 for(int j=1;j<=n2;++j){ // insert dp[0][j]=j; } for(int i=1;i<=n1;++i){ // del dp[i][0]=i; } for(int i=1;i<=n1;++i){ for(int j=1;j<=n2;++j){ if(word1[i-1]==word2[j-1]){ int temp=min(dp[i-1][j]+1,dp[i][j-1]+1); dp[i][j]=min(temp,dp[i-1][j-1]); }else{ int temp=min(dp[i-1][j]+1,dp[i][j-1]+1); dp[i][j]=min(temp,dp[i-1][j-1]+1); } } } return dp[n1][n2]; } };