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
用m*n的矩阵vector中的（i，j）存储从word1中(0...i)到word2中（0...j）的最小变化次数
矩阵中，从上到下对应删除操作，从左到右对应插入操作，从左上到右下对应Replace操作。
class Solution {
public:
int minDistance(string word1, string word2) { //1->2
if(word1==word2)
return ;
int len1 = word1.size();
int len2 = word2.size();
vector<int> temp(len2+,);
vector<vector<int> > vec(len1+,temp);
for(int i=;i<=len1;i++){
vec[i][] = i;
}
for(int j=;j<=len2;j++){
vec[][j] = j;
}
for(int i=;i<=len1;i++){
for(int j=;j<=len2;j++){
int tem1,tem2;
tem1 = min(vec[i-][j]+,vec[i][j-]+);
tem2 = word1[i-]==word2[j-] ? vec[i-][j-]:vec[i-][j-]+;
vec[i][j] = min(tem1,tem2);
}//end for
}//end for
return vec[len1][len2];
}
};