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• # LeetCode-Distinct Subsequences

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`.

该题实际上是要求S到T只通过删除操作进行变换的方法数(通过从S中删去几个字符变成T的方法)

通过动归，对S前i位与T前j位的情况 考虑均新增一位会如何变化

```class Solution {
public:
int numDistinct(string S, string T) {
// Note: The Solution object is instantiated only once and is reused by each test case.
vector<vector<int> > count;
if(S.length()<T.length())return 0;
count.resize(S.length()+1);
for(int i=0;i<=S.length();i++){
count[i].resize( min(i,(int)T.length()) +1);
count[i][0]=1;
}
for(int i=1;i<=S.length();i++)
{
if(count[i].size()==i+1){
int j=1;
for(;j<count[i].size()-1;j++){
if(S[i-1]==T[j-1]){
count[i][j]=count[i-1][j-1]+count[i-1][j];
}
else{
count[i][j]=count[i-1][j];
}
}
if(S[i-1]==T[j-1]){
count[i][j]=count[i-1][j-1];
}
else{
count[i][j]=0;
}
}
else{
int j=1;
for(;j<count[i].size();j++){
if(S[i-1]==T[j-1]){
count[i][j]=count[i-1][j-1]+count[i-1][j];
}
else{
count[i][j]=count[i-1][j];
}
}
}
}
return count[S.length()][T.length()];
}
};```
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• 原文地址：https://www.cnblogs.com/superzrx/p/3358338.html