problems72_edit-distance

1.列出狀態轉移方程：

• dp[i][j]表示：使用word1中前i個字符構造word2中前j個字符的最小操作數；
• 當word1[i] == word2[j]:dp[i][j] = dp[i-1][j-1];
• 當word1[i] != word2[j]:dp[i][j] = min({dp[i-1][j-1] (替換),dp[i-1][j] (刪除),dp[i][j-1] (插入)})+1。

2.編碼：

class Solution {
    public:
        int minDistance(string word1, string word2) {
            int n1 = word1.size