LeetCode: Edit Distance of Two Words

最新推荐文章于 2019-01-13 06:58:13 发布

原创最新推荐文章于 2019-01-13 06:58:13 发布 · 590 阅读

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#leetcode #算法 #Dynamic Programming

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本文介绍了解决两个字符串之间编辑距离问题的标准算法。通过使用动态规划方法，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步骤，包括插入、删除和替换字符等操作。

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Edit Distance of the Two Words
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
I remembered there is standard algorithm to solve this problem, discussed in courses like Natural Language Processing (see www.coursera.com)
Dynamic Programming will be used to solve this problem.
Basic Idea:
If we know the distance between word2[i-1] & word1[j-1], between word2[i] & word1[j-1], and between word2[i-1] & word1[j], then the distance between word2[i] & word1[j] will be
if(word2[i] == word1[j])
dist[i][j] = min(dist[i-1][j-1], min(dist[i-1][j], dist[i][j-1])+1)
else
dist[i][j] = min(dist[i-1][j-1], min(dist[i-1][j], dist[i][j-1])) + 1
Un-optimized code:
class Solution {
public:
    int minDistance(string word1, string word2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(word2.length() == 0)
            return word1.length();
        if(word1.length() == 0)
            return word2.length();
        int word1Length = word1.length(), word2Length = word2.length();
        int ** matrix = new int *[word2Length];
        for(int i=0; i<word2Length; i++){
            matrix[i] = new int[word1Length];
        }
        if(word1[0] == word2[0])
            matrix[0][0] = 0;
        else
            matrix[0][0] = 1;
        for(int j=1; j<word1Length; j++){
            if(word1[j] == word2[0] && j<matrix[0][j-1] + 1)
                matrix[0][j] = j;
            else
                matrix[0][j] = matrix[0][j-1]+1;
        }
        for(int i=1; i<word2Length; i++){
            if(word2[i] == word1[0] && i<matrix[i-1][0] + 1)
                matrix[i][0] = i;
            else
                matrix[i][0] = matrix[i-1][0] + 1;
            for(int j=1; j<word1Length; j++){
                int temp = min(min(matrix[i-1][j], matrix[i][j-1]), matrix[i-1][j-1]);
                matrix[i][j] = temp + 1;
                if(word1[j] == word2[i] && matrix[i-1][j-1] < temp + 1)
                    matrix[i][j] = matrix[i-1][j-1];
            }
        }
        int ret = matrix[word2Length-1][word1Length-1];
        for(int i=0; i<word2Length; i++)
            delete[] matrix[i];
        delete[] matrix;
        return ret;
    }
};