Distinct Subsequences 解题报告

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题目：给两个字符串S和T，判断T在S中出现的次数。

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

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如果当前字符相同，结果加上S和T在该index之后的匹配方法数

如果当前字符不同，将S的指针向后移，递归计算

class Solution {
private:
    int cnt;
    int len_s;
    int len_t;
public:
    Solution():cnt(0){}
    void Count(string S,string T, int idx_ss, int idx_ts){
        if(idx_ts == len_t){
            cnt++;
            return;
        }
        int i,j,k;
        for (i=idx_ss; i<len_s; i++) {
            if (S[i] == T[idx_ts]) {
                Count(S, T, i + 1, idx_ts + 1);
            }
        }
    }
    
    int numDistinct(string S, string T) {
        len_s = S.length();
        len_t = T.length();
        Count(S, T, 0, 0);
        return cnt;
    }
};

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思路2：DP

如果当前字符相同，dp[i][j]结果等于用S[i](dp[i-1][j-1])和不用S[i](dp[i-1][j])方法数求和

如果当前字符不同，dp[i][j] = dp[i-1][j]

class Solution {
private:
    int len_s;
    int len_t;
public:
    int Count(string S,string T){
        int i,j;
        int dp[len_s][len_t];
        memset(dp, 0, sizeof(dp));
        
        if (S[0]==T[0]) {
            dp[0][0] = 1;
        }
        
        for(i=1;i<len_s;i++){
            dp[i][0] = dp[i-1][0];
            if (T[0]==S[i]) {
                dp[i][0]++;
            }
        }
                
        for (i=1; i<len_s; i++) {
            for (j=1; j<len_t && j<=i; j++) {
                if (S[i]!=T[j]) {
                    dp[i][j] = dp[i-1][j];
                    //cout<<dp[i-1][j]<<endl;
                }
                else{
                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];
                    //dp[i-1][j-1]: use S[i], as S[i]==T[j]
                    //dp[i-1][j]  : don‘t use S[i]
                    //cout<<dp[i][j]<<endl;
                }
            }
        }
        return dp[len_s-1][len_t-1];
    }
    
    int numDistinct(string S, string T) {
        len_s = S.length();
        len_t = T.length();
        return Count(S, T);
    }
};

Distinct Subsequences 解题报告

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原文地址：http://blog.csdn.net/abcjennifer/article/details/40263761