LeetCode-72：Edit Distance (编辑距离)

题目：

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

例子

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

问题解析：

求两个字符串的最小编辑距离。

链接：

• https://leetcode.com/problems/edit-distance/description/

思路标签：

算法：动态规划

解答：

• 使用dp[i][j]用来表示字符串1的0~i-1、字符串2的0~j-1的最小编辑距离；
• 我们可以知道边界情况：dp[i][0] = i、dp[0][j]=j；
• 同时对于两个字符串的子串，都能分为最后一个字符相等或者不等的情况：
• 如果words[i-1] == words[j-1]：dp[i][j] = dp[i-1][j-1]；也就是说当前的编辑距离和位置i和j的字符无关；
• 如果words[i-1] != words[j-1]：则存在三种可能的操作：
  
  • 向word1插入：dp[i][j] = dp[i][j-1] + 1;
  • 从word1删除：dp[i][j] = dp[i-1][j] + 1;
  • 替换word1元素：dp[i][j] = dp[i-1][j-1] + 1;

class Solution {
public:
    int minDistance(string word1, string word2) {
        int rows = word1.length();
        int cols = word2.length();
        vector<vector<int> > dp(rows+1, vector<int>(cols+1, 0));

        for(int i=1; i<=rows; ++i)
            dp[i][0] = i;
        for(int j=1; j<=cols; ++j)
            dp[0][j] = j;

        for(int i=1; i<=rows; ++i){
            for(int j=1; j<=cols; ++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], dp[i][j-1])) + 1;
            }
        }

        return dp[rows][cols];
    }
};