edit distance leetcode

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()];
    }
};