[代码随想录一刷] day55 动态规划
392.判断子序列
class Solution {
public:
bool isSubsequence(string s, string t) {
vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1, 0));
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] + 1;
else dp[i][j] = dp[i][j - 1];
}
}
if (dp[s.size()][t.size()] == s.size()) return true;
return false;
}
};
115.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));
for (int i = 0; i < s.size(); i++) dp[i][0] = 1;
for (int j = 1; j < t.size(); j++) dp[0][j] = 0;
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()];
}
};
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