给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

插入一个字符
删除一个字符
替换一个字符
示例 1:

输入: word1 = "horse", word2 = "ros"
输出: 3
解释: 
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:

输入: word1 = "intention", word2 = "execution"
输出: 5
解释: 
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')

 

  动态规划二维数组存储 
class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> ans(word1.size()+1,vector<int>(word2.size()+1));
        for(int i=0;i<=word2.size();i++)ans[0][i]=i;
        for(int i=0;i<=word1.size();i++)ans[i][0]=i;
        for(int i=1;i<=word1.size();i++){
            for(int j=1;j<=word2.size();j++){
                ans[i][j]=min(ans[i][j-1],ans[i-1][j])+1;
                ans[i][j]=min(ans[i][j],ans[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1));
            }
        }return ans[word1.size()][word2.size()];
    }
};
动态规划，优化为一维数组存储  
class Solution {
public:
    int minDistance(string word1, string word2) {
        // vector<vector<int>> ans(word1.size()+1,vector<int>(word2.size()+1));
        // for(int i=0;i<=word2.size();i++)ans[0][i]=i;
        // for(int i=0;i<=word1.size();i++)ans[i][0]=i;
        vector<int>vv(word2.size()+1);
        for(int i=0;i<=word2.size();i++)vv[i]=i;

        int pre=0;
        for(int i=1;i<=word1.size();i++){
            vv[0]=i;
            pre=i-1;
            for(int j=1;j<=word2.size();j++){
                int t=vv[j];
                // ans[i][j]=min(ans[i][j-1],ans[i-1][j])+1;
                // ans[i][j]=min(ans[i][j],ans[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1));
                vv[j]=min(vv[j-1],vv[j])+1;
                vv[j]=min(vv[j],pre+(word1[i-1]==word2[j-1]?0:1));
                if(word1[i-1]==word2[j-1])cout<<"["<<i-1<<","<<j-1<<pre<<"]";
                cout<<vv[j];
                pre=t;
            }cout<<endl;
        }return vv[word2.size()];
    }
};