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')

dp[i][j]代表word1中前i个字符串变化为word2中前j个字符串所需最小步骤
class Solution {
    public int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        for(int i = 0; i < word1.length() + 1; i++) {
            dp[i][0] = i;
        }
        for(int i = 0; i < word2.length() + 1; i++) {
            dp[0][i] = i;
        }

        for(int i = 1; i < word1.length() + 1; i++) {
            for(int j = 1; j < word2.length() + 1; j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1)) dp[i][j] = dp[i-1][j-1];
                //                     第i字符与第j字符相同，那就只用考虑1和2去掉这个字符前面的情况
                else dp[i][j] = Math.min(Math.min(dp[i-1][j-1]+1, dp[i-1][j]+1), dp[i][j-1]+1);
                // 第i字符与第j字符不同  1的第i个字符替换成2的第j个字符  1的第i个字符删掉 1添加2的第j个字符
            }
        }

        return dp[word1.length()][word2.length()];
    }
}