115. Distinct Subsequences
题目
Given a string S and a string T, count the number of distinct subsequences of S which equals T.
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.
解析
- 此题花费很多时间，对递推公式理解不清楚，用一维表示减少空间
- 对比最大公共子序列和子串
//115. Distinct Subsequences
class Solution_115 {
public:
int numDistinct(string S, string T) { //母串和子串匹配的次数
int lenx = T.size(); //子串
int leny = S.size(); //母串
if (lenx==0||leny==0)
{
return 0;
}
vector<vector<int> > dp(leny + 1, vector<int>(lenx + 1, 0));
for (int i = 0; i <= leny;i++) //遍历母串
{
for (int j = 0; j <= lenx;j++) //遍历子串
{
if (j==0)
{
dp[i][j] = 1; //当子串长度为0时，所有次数都是1
continue;
}
if (i>=1&&j>=1)
{
if (S[i - 1] == T[j - 1]) //当前母串和子串当前元素相等
{
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
}
else
{
dp[i][j] = dp[i-1][j];
}
}
}
}
return dp[leny][lenx];
}
int numDistinct2(string s, string t) {
vector<int> match(t.size() + 1);
match[0] = 1;
for (int i = 0; i<s.size(); i++)
for (int j = t.size(); j>0; j--)
match[j] += (t[j - 1] == s[i] ? match[j - 1] : 0);
return match[t.size()];
}
int numDistinct1(string S, string T) { //bug.计算最长公共子序列
//测试用例:
// "ddd", "dd"
// 对应输出应该为:3
// 你的输出为 : 2
int lenx = S.size();
int leny = T.size();
if (lenx==0||leny==0)
{
return 0;
}
vector<vector<int>> vecs(leny+1, vector<int>(lenx+1, 0));
for (int i = 1; i <= T.size();i++) //行
{
for (int j = 1; j <=lenx;j++) //列
{
if (S[j-1]==T[i-1])
{
vecs[i][j] = vecs[i-1][j - 1] + 1;
}
else
{
vecs[i][j] = max(vecs[i - 1][j], vecs[i][j - 1]);
}
}
}
int cnt = 0;
if (vecs[leny][lenx] > 0){
cnt++;
for (int i = lenx - 1; i > 0; i--)
{
if (vecs[leny][i] == vecs[leny][lenx])
{
cnt++;
}
}
}
return cnt;
}
};
题目来源
