Level：
Hard
题目描述：
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')思路分析：
动态规划的思想，dp[i] [j]代表将word1前i个字符转换成word2前j个字符所需要的操作次数。
如果word1[i]==word2[j],那么dp[i] [j]=dp[i-1] [j-1]。
如果word1[i]不等于word2[j]，需要找出插入元素，删除元素，替换元素中最小的操作，然后加一。
dp[i] [j-1]表示word1前i个可以表示word2前j-1个，那么要表示前j个，只能执行插入操作。
dp[i-1] [j]表示word1前i-1个可以表示word2前j个，那么要前i个表示前j个，只能执行删除操作。
dp[i-1] [j-1]表示word1前i-1个可以表示word2前j-1个，那么要前i个表示前j个，只能执行替换操作。
则dp[i] [j]=min(dp[i-1] [j],dp[i] [j],dp[i-1] [j-1])+1;
代码：
public class Solution{
public int minDistance(String word1,String word2){
int m=word1.length();
int n=word2.length();
int [][]dp=new int [m+1][n+1];
for(int i=0;i<=m;i++){
dp[i][0]=i; //单纯的删除操作
}
for(int i=0;i<=n;i++){
dp[0][i]=i; //单纯的插入操作
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(word1.charAt(i)==word2.charAt(j)){
dp[i+1][j+1]=dp[i][j];
}else{
dp[i+1][j+1]=Math.min(dp[i+1][j],Math.min(dp[i][j+1],dp[i][j]))+1;
}
}
}
return dp[m][n];
}
}