leetcode : Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

求解方法类似于LCS问题，用dp[i][j]表示word1的前i个字母 到word2的前j个字母的步骤。

状态转移公式:

　　　if(word1[i - 1] == word2[j - 1]

　　　　　　dp[i][j] = dp[i - 1][j - 1]

　　　else

　　　　　　dp[i][j] = min{dp[i-1][j],  dp[i][j - 1],  dp[i - 1][j - 1]} + 1

AC代码:

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 = 0; i < word1.size() + 1; ++i){
           dp[i][0] = i;
       }
       for(int j = 0; j < word2.size() + 1; ++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], min(dp[i - 1][j], dp[i][j - 1])) + 1;
               }
           }
       }
       return dp[word1.size()][word2.size()];
    }
};