leetcode115. Distinct Subsequences-优快云博客

本文链接：https://blog.youkuaiyun.com/barebear007/article/details/79488796

本文介绍了一种用于计算两个字符串中一个作为另一个子序列出现次数的算法。该算法通过动态规划实现，能够有效地解决字符串匹配问题。

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class Solution {
public:
int numDistinct(string s, string t) {
if(t.size()<=0) return 1;
if((s.size()<=0 && t.size()>0) || s.size()<t.size()){
return 0;
}
int ssize = s.size();
int tsize = t.size();
vector<vector <int> > matchs(tsize ,vector<int>(ssize,0));
int tcursor = tsize - 1;
int sStart = tcursor;
int sEnd = tcursor - tsize + ssize;
int scursor = sEnd;
while(tcursor>=0){
sStart = tcursor;
sEnd = tcursor - tsize + ssize;
scursor = sEnd;
while(scursor>=sStart){
//cout<<"111 "<<matchs[tcursor][scursor]<<" "<< tcursor<<" "<<scursor<< " "<<sStart<<" "<<ssize<<endl;


if(t[tcursor] == s[scursor]){
if(scursor>=ssize - 1 && tcursor>=tsize - 1)
matchs[tcursor][scursor] += 1;
else if(tcursor>=tsize - 1){
matchs[tcursor][scursor] += 1+matchs[tcursor][scursor+1];
}
else{
matchs[tcursor][scursor] = matchs[tcursor+1][scursor+1] + matchs[tcursor][scursor+1];
//cout<<matchs[tcursor][scursor]<<" "<< tcursor<<" "<<scursor<< " "<<sStart<<" "<<ssize<<endl;
}
}else{
if(scursor>=ssize - 1 && tcursor>=tsize - 1)
matchs[tcursor][scursor] =0;
else
matchs[tcursor][scursor] = matchs[tcursor][scursor+1];
}
scursor--;
}

tcursor--;
}
return matchs[0][0];
}