72. Edit Distance 编辑距离

最新推荐文章于 2024-04-22 19:06:35 发布

转载最新推荐文章于 2024-04-22 19:06:35 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/xiejunzhao/p/8306154.html

本文介绍了一种计算两个字符串之间编辑距离的算法，并提供了详细的实现步骤。编辑距离是指通过插入、删除或替换字符的方式将一个字符串转换成另一个字符串所需的最少操作次数。

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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

给定两个单词word1和word2，找到将word1转换为word2所需的最小步数。（每个操作都算作1步。）

你有一个单词允许以下3个操作：

a）插入一个字符

b）删除一个字符

c）替换一个字符

/** * @param {string} word1 * @param {string} word2 * @return {number} */var minDistance = (word1, word2) => { let lenA = word1.length; let lenB = word2.length; let d = createMatrix(lenB + 1, lenA + 1); for (let i = 0; i <= lenA; i++) { d[i][0] = i; } for (let i = 0; i <= lenB; i++) { d[0][i] = i; } for (let i = 1; i <= lenA; i++) { for (let j = 1; j <= lenB; j++) { if (word1[i - 1] == word2[j - 1]) { d[i][j] = d[i - 1][j - 1]; } else { d[i][j] = Math.min(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1] + 1); } } } return d[lenA][lenB];}let createMatrix = (rowNum, colNum) => { let matrix = []; matrix.length = colNum; for (let i = 0; i < matrix.length; i++) { matrix[i] = []; matrix[i].length = rowNum; matrix[i].fill(0); } return matrix}let a = "abcde";let b = "abcd";console.log(minDistance(a, b));

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转载于:https://www.cnblogs.com/xiejunzhao/p/8306154.html