[刷题笔记55 动态规划15]

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目录

• 动态规划
  • 392.判断子序列
  • 115.不同的子序列

动态规划

● 392.判断子序列
● 115.不同的子序列

392.判断子序列

392.判断子序列

法1：动态规划

    bool isSubsequence(string s, string t) {
        //动态规划
        vector<vector<int>>dp(s.size() + 1,vector<int>(t.size() + 1, 0));
        for (int i = 1; i <= s.length(); ++i) {
            for (int j = 1; j <= t.length(); ++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.length())return true;
        else return false;
    }

//法2：双指针

    bool isSubsequence(string s, string t) {
        //双指针
        int m =s.length(); int n =t.length();
        int l = 0,r = 0;
        while (l < m && r < n){
            if (s[l] == t[r]){
                l++;
            }
            r++;
        }
        return (l == m);
}

115.不同的子序列

115.不同的子序列
法1：动态规划

    int numDistinct(string s, string t) {
        vector<vector<unsigned int>>dp(s.size() + 1,vector<unsigned int>(t.size() + 1, 0));
        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.length()][t.length()];
    }