edit distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

class Solution {public:    int minDistance(string word1, string word2) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int result = 0;         int n1 = word1.size();          int n2 = word2.size();          int dist[n1 + 1][n2 + 1];          memset(dist, 0, sizeof(int) * (n1 + 1) * (n2+1));                 for (int i = 0; i < n1+1; i++)            dist[i][0] = i;                    for (int i = 0; i < n2+1; i++)            dist [0][i] = i;                    for (int i = 1; i< n1+1; i++)            for (int j = 1; j < n2+1; j++){                if (word1[i-1] == word2[j-1])                    dist[i][j] = dist [i-1][j-1];                else{                    dist[i][j] = min(dist [i][j-1], dist [i-1][j])+1;                    dist[i][j] = min(dist[i][j], dist[i-1][j-1]+1);                }              }          return dist[n1][n2];    }};

这一题与longest subsequnce相关，找到最长子序列也就知道了需要的操作次数。

1. 原始的思想史：找到最长子序列的长度，用word1的长度减去最长子序列的长度即可。但是这样做有一个不方便：如果三个操作的权重不同该怎么办？
2. 仍然用DP，但是dp_value就是最小的操作。
3. 

dist[i][j] = min(dist [i][j-1], dist [i-1][j])+1;dist[i][j] = min(dist[i][j], dist[i-1][j-1]+1);

原本没有考虑第二行，导致错误！！！

4. 注意新的dp生成及初始化方法，之前都是用vector