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

分析：假設dp[i][j]表示word1[0…i]和word2[0…j]的最小編輯距離。
如果word1為空，則刪除word2中元素
如果word2為空，則刪除word1中元素
都不為空，有三種選擇：
1）刪除word1的最後一個字元，此時dp[i-1][j] + 1
2) 刪除word2的最後一個字元，此時dp[i][j-1] + 1
3) word1最後一個替換為word2最後一個 , 則dp[i-1][j-1] + (word1[i] != word2[j])(下標從1開始)

程式碼如下：

class Solution {
public:
    int minDistance(string word1,string word2)
    {
        int m = word1.length();
        int n = word2.length();
        int dp[m+1][n+1];
        memset(dp,0,sizeof(dp));

        for(int i = 0; i <= m; i++) dp[i][0] = i;
        for(int i = 0; i <= n; i++) dp[0
 
][i] = i;

        for(int i = 1; i <= m; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                int num ;
                if(word1[i-1] == word2[j-1]) num = 0;
                else num = 1;

                dp[i][j] = min(dp[i-1][j] + 1, min(dp[i][j-1] + 1, dp[i-1
 
][j-1] + num) );
            }
        }
        return dp[m][n];
    }
};