这样的字符转换的dp挺经典的，
若word1[i+1]==word2[j+1] dp[i+1][j+1] = dp[i][j]；否则，dp[i+1][j+1] = dp[i][j] + 1。（替换原则），dp[i+1][j+1]还能够取dp[i][j+1]与dp[i+1][j]中的较小值。（删除加入原则）
class Solution {
public:
int minDistance(string word1, string word2) {
int dp[word1.size()+1][word2.size()+1];
for(int i = 0; i < word1.size()+1; i++)
dp[i][0] = i;
for(int i = 0; i < word2.size()+1; i++)
dp[0][i] = i;
for(int i = 0; i < word1.size(); i++) {
for(int j = 0; j < word2.size(); j++) {
if(word1[i] == word2[j])
dp[i+1][j+1] = dp[i][j];
else
dp[i+1][j+1] = min(min(dp[i][j],dp[i][j+1]),dp[i+1][j])+1;
}
}
return dp[word1.size()][word2.size()];
}
};
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5222739.html，如需转载请自行联系原作者
