Edit Distance

Posted 2020-08-09 wxquare的学习笔记

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) {
        int l1 = word1.length();
        int l2 = word2.length();

        vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1, 0));
        for (int i = 1; i <= l1; i++) {
            dp[i][0] = i;
        }
        for (int i = 1; i <= l2; i++) {
            dp[0][i] = i;
        }

        for (int i = 1; i <= l1; i++) {
            for (int j = 1; j <= l2; 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[l1][l2];
    }
};

class Solution2 {
public:
    int minDistance(string word1, string word2) {
        int l1 = word1.length();
        int l2 = word2.length();

        int dp[l2 + 1] { 0 };
        for (int i = 1; i <= l2; i++) {
            dp[i] = i;
        }

        for (int i = 1; i <= l1; i++) {
            int pre = i;
            for (int j = 1; j <= l2; j++) {
                int cur;
                if(word1[i-1] == word2[j-1]){
                    cur = dp[j-1];
                }else{
                    cur = min(dp[j-1],min(dp[j],pre))+1;
                }
                dp[j-1] = pre;
                pre = cur;
            }
            dp[l2] = pre;
        }
        return dp[l2];
    }
};