代码随想录算法训练营Day54 ||leetCode 392.判断子序列 || 115.不同的子序列

392.判断子序列 

双指针遍历，比较简单易懂

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int a = 0, b = 0;
        while(a < s.size() && b < t.size()){
            if (s[a] == t[b]){
                a++;
            }
            b++;
        }
        return a == s.size();
    }
};

115.不同的子序列

这道题初始化与状态方程都不好想，建议多看解析

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()];
    }
};