【LeetCode】Distinct Subsequences

参考链接

http://blog.csdn.net/xshalk/article/details/8223120

http://blog.csdn.net/bigapplestar/article/details/17523391

http://blog.csdn.net/jellyyin/article/details/9060709

题目描述

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

.

题目分析

题意：

给定2个字符串a, b，求b在a中出现的次数。要求可以是不连续的，但是b在a中的顺序必须和b以前的一致。
示例说明：
S = "rabbbit", T = "rabbit"

为了区分3个b，我们写作b0b1b3
T在S中出现了3分。分别是rab0b1it，rab0b2it，rab1b2it

思路：动态规划的关键就是找递归公式。

类似于数字分解的题目。dp[i][j]表示：b的前j个字符在a的前i个字符中出现的次数。

如果a[i] == b[j]则 dp[i][j] = dp[i-1][j] + dp[i-1][j-1]

如果a[i] != b[j]则 dp[i][j] = dp[i-1][j]

总结

这个一个典型的动态规划题，需要多多复习。

代码示例

#include
#include
using namespace std;
void printvec(vector A)
{
for(int i = 0;i > A, char *name)
{
printf("%s\n",name);
for(int i = 0;i > num(S.size()+1,vector(T.size()+1));
for(int i = 0; i<=S.size(); i++){
num[i][0] = 0;
}
for(int j=0; j<=T.size(); j++){
num[0][j] = 0;
}

for(int i = 1; i<=S.size(); i++){
for(int j = 1; j<=i && j<=T.size(); j++){
num[i][j] = num[i-1][j];
if(S[i-1] == T[j-1]){
if(j == 1){
num[i][j] += 1;
}else{
num[i][j] += num[i-1][j-1];
}
}
}
}

//printvecvec(num,"ret");
return num[S.size()][T.size()];
}
};
#elif 1
class Solution {
public:

int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int len_s = S.size();
int len_t = T.size();
vector  > f(len_s+1,vector(len_t+1));
if(len_s < len_t)
return 0;
for(int i = 0 ; i <= len_s ; i++)
f[i][0] = 0;
for(int j = 0 ; j <= len_t ; j++)
f[0][j] = 0;

for(int i = 1 ; i <= len_s ;i++)
if(S[i-1]==T[0])
f[i][1] = f[i-1][1] + 1;
else
f[i][1] = f[i-1][1];
for(int i = 2;  i <= len_s ; i++)
for(int j = 2; j <= len_t ; j++)
if(S[i-1]==T[j-1])
f[i][j] = f[i-1][j-1] + f[i-1][j];
else
f[i][j] = f[i-1][j];
return f[len_s][len_t];
}
};
#endif
void test0()
{
Solution so;
if(so.numDistinct("rabbbit","rabbit") != 3)
printf("------------------------failed\n");
else
printf("------------------------passed\n");
}
int main(int argc, char *argv[])
{
test0();
return 0;
}

推荐学习C++的资料

C++标准函数库
http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询
http://www.cplusplus.com/