LeetCode开心刷题三十一天——72. Edit Distance

72. Edit Distance

Hard

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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:

1. Insert a character
2. Delete a character
3. 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;
}