LeetCode OJ-- Edit Distance **

https://oj.leetcode.com/problems/edit-distance/

动态规划，它的小规模问题是：当 word1  word2都比较小的时候，word1变成 word2需要的变化数。

设数组 record[i][j] 表示word1 从0到 i ，word2 从 0 到 j 的 word1 到 word2 需要经过几次变化。

列等式如下：

record[i][j] = record[i-1][j-1]   if 'i'=='j'

record[i][j] = record[i-1][j-1] +1  replace                     if 'i'!='j'

                  record[i][j-1] +1     delete

                 record[i-1][j] +1      insert

取它们3个的最小值。

初始化: record[0][j] = j

         record[i][0]  = i

class Solution {
public:
    int minDistance(string word1, string word2) {
        //word1 is smaller
        if(word1.size()>word2.size())
        {
            string temp;
            temp = word1; word1 = word2; word2 = temp;
        }
        if(word1.empty())
            return word2.size();
        if(word1 == word2)
            return ;

        //find the max same bits
        int maxSame = sameBits(word1,word2);

        return maxSame;
    }
    int sameBits(string &word1,string &word2)
    {
        vector<vector<int> > record;
        record.resize(word1.size()+);
        for(int i = ; i<= word1.size();i++)
            record[i].resize(word2.size()+);

        for(int i = ;i<=word1.size();i++)
            record[i][] = i;
        for(int j = ;j<=word2.size();j++)
            record[][j] = j;

        for(int row = ;row<= word1.size();row++)
            for(int col = ;col<=word2.size();col++)
            {
                if(word1[row-] == word2[col-])
                    record[row][col] = record[row-][col-];
                else
                {
                    int _min = min(record[row-][col],record[row][col-]);
                    record[row][col] = min(_min,record[row-][col-]) + ;
                }
            }

        return record[word1.size()][word2.size()];
    }
};
int main()
{
    class Solution mys;
    string str1 = "distance";
    string str2 = "springbok";
    cout<<mys.minDistance(str1,str2);
}

可以 const size_t lenword1 = word1.size()

      const size_t lenword2 = word2.size()

int  record[lenword1+1][lenword2+1];

这样定义数组