Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
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')
This question itself has a clear mind.
First,divide into two main situation d1[i]==d2[j],
if equal means don't need add distance,if not add 1
except for all of this,each time you see it you need compare dp[i][j-1],dp[i-1][j],dp[i-1][j-1]
select a minimum one.and don't forget to add extra distance.It's important to consider the whole
picture.
ATTENTION:
Judge NULL situation can largely decrease the running time and bigger a bit space(NIPTTnot important)
#include<stdio.h> #include<iostream> #include<string> #include<vector> #include<set> #include<map> #include<algorithm> using namespace std; class Solution { public: int minDistance(string word1, string word2) { if(word1.empty()||word2.empty()) {int tmp=word1.length()-word2.length(); return abs(tmp);} //==situation int m=word1.length(),n=word2.length();int tmp=0; vector<vector<int>> dp= vector<vector<int>>(m + 1, vector<int>(n + 1, 0)); for(int i=0;i<m+1;i++) dp[i][0]=i; for(int j=0;j<n+1;j++) dp[0][j]=j; for(int i=1;i<m+1;i++) { for(int j=1;j<n+1;j++) { int c=(word1[i-1]==word2[j-1])?0:1; dp[i][j]=min(min(dp[i-1][j],dp[i][j-1])+1,dp[i-1][j-1]+c); } } return dp[m][n]; } }; int main() { Solution s; //string word1="intention";string word2 = "execution"; string word1 = "horse", word2 = "ros"; //string s1=" "; int res=s.minDistance(word1,word2); cout<<res<<endl; return 0; }