72. Edit Distance-优快云博客

本文链接：https://blog.youkuaiyun.com/dreamjay1997/article/details/87862751

本文深入探讨了编辑距离算法，一种用于衡量两个字符串相似度的方法。通过插入、删除或替换字符的操作，文章详细解释了如何计算从一个单词转换到另一个单词所需的最小操作数，并提供了具体的例子进行说明。

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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

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')

最近在做关于文本相似度的项目，edit distance是判断文本相似度的一种算法

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
    let len1 = word1.length
    let len2 = word2.length
    let matrix = [], temp = 0 
    for(let i = 0; i <= len1; i ++){ 
        matrix[i] = []
        matrix[i][0] = i
    }
    for(let i = 1; i <= len2; i ++) {
        matrix[0][i] = i
    }
    for(let i = 1; i <= len1; i ++) {
        for(let j = 1;j <= len2; j ++) {
            word1[i - 1] === word2[j - 1] ? temp = 0 : temp = 1 
            matrix[i][j] = Math.min(matrix[i - 1][j] + 1, matrix[i][j - 1] + 1, matrix[i - 1][j - 1] + temp)
        }
    }   
    return matrix[len1][len2]
    
};

Runtime: 112 ms, faster than 50.00% of JavaScript online submissions for Edit Distance.

Memory Usage: 41.5 MB, less than 43.59% of JavaScript online submissions for Edit Distance.