代码随想录算法训练营第五十四天|392.判断子序列 115.不同的子序列

392.判断子序列

思路

代码

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;
        else return false;
    }
};

总结

115.不同的子序列

思路

代码

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()];

    }
};

总结