经典的dp题
const int MAXN = 1000;
int dp[MAXN][MAXN];
class Solution {
public:
int minDistance(string word1, string word2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
for(int i = 0; i <= word1.size(); ++ i)
dp[i][0] = i;
for(int i = 0; i <= word2.size(); ++ i)
dp[0][i] = i;
for(int i = 1; i <= word1.size(); ++ i)
for(int j = 1; j <= word2.size(); ++ j){
if(word1[i - 1] == word2[j - 1])
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
}
return dp[word1.size()][word2.size()];
}
};