leetcode 115: Distinct Subsequences

最新推荐文章于 2022-03-13 16:21:53 发布

mikefan1991

最新推荐文章于 2022-03-13 16:21:53 发布

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本文介绍如何使用动态规划解决子串匹配问题，详细解释了dp数组的定义及更新规则，通过实例演示了算法的具体实现过程。

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Use DP, dp[i][j] means the number of substrings of t before index i in the substring of s before index j. When I loop the dp array, I have two conditions that t[i]==s[j] and t[i]!=s[j]. If t[i]==s[j], this s[j] can be considered or ignored, which means dp[i][j]=dp[i-1][j-1]+dp[i][j-1]. If t[i]!=s[j], this s[j] must be ignored, so dp[i][j]=dp[i][j-1].

class Solution {
public:
    int numDistinct(string s, string t) {
        int slen=s.length();
        int tlen=t.length();
        if(tlen>slen)
            return 0;
        vector<int> a(slen,0);
        vector<vector<int> > dp(tlen,a);
        dp[0][0]=s[0]==t[0];
        for(int i=1;i<slen;i++)
            dp[0][i]=s[i]==t[0]?dp[0][i-1]+1:dp[0][i-1];
        for(int i=1;i<tlen;i++)
            for(int j=1;j<slen;j++)
            {
                if(t[i]==s[j])
                    dp[i][j]=dp[i-1][j-1]+dp[i][j-1];//s[j] is either considered or ignored
                else
                    dp[i][j]=dp[i][j-1];//s[j] is ignored
            }
        return dp[tlen-1][slen-1];
    }
};