Distinct Subsequences

最新推荐文章于 2021-03-20 14:23:47 发布

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本文介绍了一种计算两个字符串间不同子序列数量的方法。通过一个具体的例子，即在字符串S中寻找字符串T的所有可能子序列出现次数，展示了一种有效算法的实现细节。该算法使用动态规划思想，能够高效地解决问题。

Distinct SubsequencesOct 19 '123689 / 10619

Given a string S and a string T, count the number of distinct subsequences of T in S.

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.

最长公共子串的变形。大体意思是字串T在S中有几个？

class Solution {
public:
    int numDistinct(string S, string T) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int lenS = S.size();
        if(lenS == 0)
            return 0;
        int lenT = T.size();
        vector<int> ref(lenS, 1);
        vector<int> cur(lenS, 0);
        for(int i = 0; i < lenT; i ++)
        {
            for(int j = 0; j < lenS; j++)
                cur[j] = 0;
            for(int j = i; j < lenS; j ++)
            {
                if(T[i] == S[j])
                {
                    if(j == 0)
                        cur[j] = 1;
                    else
                        cur[j] = ref[j-1] + cur[j-1];
                }
                else
                {
                    if(j == 0)
                        cur[j] = 0;
                    else
                        cur[j] = cur[j-1];
                }
            }
            for(int j = 0; j < lenS; j++)
            {
                ref[j] = cur[j];
            }
        }
        return ref[lenS-1];
    }
};