LeetCode OJ - Edit Distance

最新推荐文章于 2020-05-29 21:53:59 发布

原创最新推荐文章于 2020-05-29 21:53:59 发布 · 698 阅读

0 ·

CC 4.0 BY-SA版权

基础篇专栏收录该内容

154 篇文章

订阅专栏

本文深入探讨了最小编辑距离算法的原理与应用，通过动态规划方法解决字符串转换问题，详细解释了状态转移方程和代码实现，适用于计算机科学领域的学习与实践。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

分析：典型动态规划，使用二维记录表，dp[i][j]表示从word1字符串1到i、word2字符串从1到j位置的最小编辑次数。

dp[i][j] = min{dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + word1[i-1] == word2[j-1] ? 0 : 1}

class Solution {
public:
    int minDistance(string word1, string word2) {
        int row = word1.size() + 1;
        int col = word2.size() + 1;
        
        vector<vector<int> > f(row, vector<int>(col));

        for (int i = 0; i < row; i++)
            f[i][0] = i;

        for (int i = 0; i < col; i++)
            f[0][i] = i;

        for (int i = 1; i < row; i++)
            for (int j = 1; j < col; j++){
                f[i][j] = min(f[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 : 1), min(f[i-1][j]+1, f[i][j-1]+1));
            }

        return f[row-1][col-1];
    }
};