Edit Distance

最新推荐文章于 2019-08-05 20:44:47 发布

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最新推荐文章于 2019-08-05 20:44:47 发布

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本文详细介绍了如何计算两个字符串之间的编辑距离，包括插入、删除和替换操作，并通过示例代码展示了具体实现过程。

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题意是求俩字符串的编辑距离，编辑定义有三种1、插入字符 2、删除字符 3、替换字符。

     int minDistance(string word1, string word2) 

    {

        if (word1.size() == 0) return (int)word2.size();

        if (word2.size() == 0) return (int)word1.size();

        int result = 0;

        int *dist = new int[(word1.size() + 1) * (word2.size() + 1)];

        for (size_t i = 0; i <= word1.size(); ++i)

        {

            dist[i] = (int)i;

        }

        for (size_t j = 0; j <= word2.size(); ++j)

        {

            dist[j * word1.size()] = (int)j;

        }

        for (size_t i = 1; i <= word1.size(); ++i)

        {

            for (size_t j = 1; j <= word2.size(); ++j)

            {

                if (word1[i - 1] == word2[j - 1])

                {

                    // dist[i, j] = dist[i - 1, j - 1]

                    dist[j * word1.size() + i] = 

                            dist[(j - 1) * word1.size() + i - 1];

                }

                else

                {

                    // minDist = min(dist[i - 1, j], dist[i, j - 1])

                    const int minDist = min(

                            dist[j * word1.size() + (i - 1)],

                            dist[(j - 1) * word1.size() + i]);

                    // dist[i, j] = min(minDist, dist[i - 1, j - 1]) + 1

                    dist[j * word1.size() + i] = min(minDist,

                            dist[(j - 1) * word1.size() + i - 1]) + 1;

                }

            }

        }

        result = dist[word2.size() * word1.size() + word1.size()];

        delete[] dist;

        return result;

    }