72. Edit Distance

[Tcdian]

72. Edit Distance

给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

插入一个字符 删除一个字符 替换一个字符

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')

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
    const dp = Array.from(new Array(word1.length + 1), () => new Array(word2.length + 1));
    for (let i = 0; i <= word1.length; i++) {
        for (let j = 0; j <= word2.length; j++) {
            if (i === 0 || j === 0) {
                dp[i][j] = i | j;
            } else if (word1[i - 1] === word2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
            }
        }
    }
    return dp[word1.length][word2.length];
};

• TypeScript Solution

var minDistance = function(word1: string, word2: string): number {
    const dp: number[][] = Array.from(new Array(word1.length + 1), () => new Array(word2.length + 1));
    for (let i = 0; i <= word1.length; i++) {
        for (let j = 0; j <= word2.length; j++) {
            if (i === 0 || j === 0) {
                dp[i][j] = i | j;
            } else if (word1[i - 1] === word2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1;
            }
        }
    }
    return dp[word1.length][word2.length];
};