[LeetCode-72] Edit Distance

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

 

DP特训第一弹~某种情况下的最小路径，就是分别增、删、改所产生结果的最小值，中间结果记录即可~

 

class Solution {
    public:
        int getMinDistance(const string& word1, int let_num1,
                const string& word2, int let_num2,                                  
                int** result_map)                                                  
        {                                                                          
            if (-1 != result_map[let_num1][let_num2])
            {
                return result_map[let_num1][let_num2];                              
            }
            int dis_d = getMinDistance(word1, let_num1 - 1, word2, let_num2, result_map) + 1;        
            int dis_i = getMinDistance(word1, let_num1, word2, let_num2 - 1, result_map) + 1;
            int dis_r = getMinDistance(word1, let_num1 - 1, word2, let_num2 - 1, result_map) +      
                (word1[let_num1 - 1] == word2[let_num2 - 1] ? 0 : 1);
            result_map[let_num1][let_num2] = min<int>(min<int>(dis_d, dis_i), dis_r);
            return result_map[let_num1][let_num2];
        }                                                                          
        int minDistance(string word1, string word2) {
            // Start typing your C/C++ solution below                              
            // DO NOT write int main() function                                    
            int len1 = word1.length();
            int len2 = word2.length();
            int** result = new int* [len1 + 1];                                        
            for (int i = 0; i <= len1; ++i)                                          
            {
                result[i] = new int [len2 + 1];                                        
                result[i][0] = i;
                for (int j = 1; j <= len2; ++j)                                      
                {                                                                  
                    result[i][j] = ((0 == i) ? j : -1);                              
                }
            }
            return getMinDistance(word1, len1, word2, len2, result);                
        }  
};