day38

不同的子序列
class Solution {
public:
int numDistinct(string s, string t) {
vector<vector<uint64_t>> dp(s.size()+1, vector<uint64_t>(t.size()+1, 0));
for(int i = 0; i < s.size(); ++i)
{
dp[i][0] = 1;
}
for(int i = 1; i <= s.size(); ++i)
{
for(int j = 1; j <= t.size(); ++j)
{
if(s[i-1] == t[j-1])
{
dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
}
else
{
dp[i][j] = dp[i-1][j];
}
}
}
return dp[s.size()][t.size()];
}
};

两个字符串的删除操作

class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector> dp(word1.size()+1, vector(word2.size() + 1, 0));
for(int i = 0; i <= word1.size(); ++i)
{
dp[i][0] = i;
}
for(int j = 0; j <= word2.size(); ++j)
{
dp[0][j] = j;
}
for(int i = 1; i <= word1.size(); ++i)
{
for(int j = 1; j <= word2.size(); ++j)
{
if(word1[i-1] == word2[j-1])
{
dp[i][j] = dp[i-1][j-1];
}
else
{
dp[i][j] = min(min(dp[i-1][j], dp[i][j-1])+1, dp[i-1][j-1] +2);
}
}
}
return dp[word1.size()][word2.size()];
}
};

编辑距离

class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector> dp(word1.size()+1, vector(word2.size()+1, 0));
for(int i = 0; i <= word1.size(); ++i)
{
dp[i][0] = i;
}
for(int j = 0; j <= word2.size(); ++j)
{
dp[0][j] = j;
}
for(int i = 1; i <= word1.size(); ++i)
{
for(int j = 1; j <= word2.size(); ++j)
{
if(word1[i-1] == word2[j-1])
{
dp[i][j] = dp[i-1][j-1];
}
else
{
dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1])+1;
}
}
}
return dp[word1.size()][word2.size()];
}
};