115. Distinct Subsequences
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/*使用动态规划来解决dp[i][j] 代表 s[0~j-1]中包含的t[0~i-1]的不同子字符串的个数dp[0][j]=1 代表空字符串是任意一个字符串的子串dp[i][0]=0 (i!=0) 任意一个非空字符串都不可能是空字符串的子集dp[i][j] if(s[j-1]!=t[i-1]) dp[i][j]=dp[i][j-1] if(s[j-1]==t[i-1]) dp[i][j]=dp[i][j-1]+dp[i-1][j-1]*/class Solution {public: int numDistinct(string s, string t) { vector<vector<int>> dp(t.length()+1,vector<int>(s.length()+1,0)); for(int i=0;i<=s.length();i++) dp[0][i]=1; for(int i=1;i<=t.length();i++) for(int j=1;j<=s.length();j++) { if(t[i-1]==s[j-1]) dp[i][j]=dp[i][j-1]+dp[i-1][j-1]; else dp[i][j]=dp[i][j-1]; } return dp[t.length()][s.length()]; }};
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