Edit Distance
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;
}
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