leetcode_72. Edit Distance

題目：

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

這道題課上講過，三種方式，替換刪除增加。dp[size1+1][size2+1]表示步數。

遍歷兩個字符串，如果當前位置相同，則dp[i][j]就等於前一位置的步數，即不增加。即dp[i][j]=dp[i-1][j-1]。

如果不相同，就有三種方式變爲相同，即插入：dp[i][j-1]+1,刪除：dp[i-1][j]+1,替換：dp[i-1][j-1]+1。

取三者的最小值。

最終dp[size1][size2]就是Edit Distance。

class Solution {
public:
    int minDistance(string word1, string word2) {
        int s1 = word1.size();
        int s2 = word2.size();
        
        int dp[s1+1][s2+1];
        
        for(int i = 0; i <= s1; ++i){
            dp[i][0] = i;
        }
        
        for(int j = 0; j <= s2; ++j){
            dp[0][j] = j;
        }
        
        for(int i = 1; i <= s1; ++i){
            for(int j = 1; j <= s2; ++j){
                if(word1[i-1] == word2[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = min(dp[i-1][j]+1, min(dp[i-1][j-1] + 1, dp[i][j-1] +1));
                }
                
            }
        }
        return dp[s1][s2];
    }
};