題目

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

題目大意就是給出兩個字串，需要經過多少步才能從word1變成word2，可以用的操作為插入，刪除，新增

分析

題目的本質就是讓你求個兩個字串之間的編輯距離，和《演算法概論》(中文版)第177頁中的編輯距離一樣
在這裡我按照我的想法大致解釋一下
假設arr[i][j]

• word1[0,i)由word1[0, i-1)插入一個字元word2[j-1]得到
• word2[0,j)由word2[0,j-1)插入一個字元word1[i-1]得到
• word1[0,i)有word1[0,i-1)插入一個word1[i-1]得到，word2[0,j)有word2[0,j-1)新增word2[j-1]得到
  以上三種情況對應的距離為
• arr[i][j] = arr[i-1][j]+1
• arr[i]]j] = arr[i][j-1] + 1
• arr[i][j] = arr[i-1][j-1] + (word[i-1]==word[j-1] ? 0 : 1)

程式碼

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();
        int arr[m+1][n+1];
        for (int i = 0; i <= m; ++i) {
            arr[i][0] = i;
        }
        for (int j = 0; j <= n; ++j){
            arr[
 
0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int min = arr[i-1][j] + 1;
                if(min > arr[i][j-1] + 1){
                    min = arr[i][j-1] + 1;
                }
                int diff = word1[i-1] == word2[j-1] ? 0 : 1;
                if(min > arr[i-1][j-1] + diff){
                    min = arr[i-1][j-1] +diff;
                }
                arr[i][j] = min;
            }
        }
        return arr[m][n];
    }
};