LeetCode: Distinct Subsequences

[b]LeetCode: Distinct Subsequences[/b]

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

.

地址：https://oj.leetcode.com/problems/distinct-subsequences/

算法：这道题的题目描述的不是很清楚。按照题目给出的例子，应该是在S中寻找等于T的子序列，然后求这样的子序列的个数。我是用动态规划解决的，用二维dp来存储子问题的解，其中dp[i][j]表示子问题(T[0~i],S[0~j])的解，这样我们就可以按行优先来完成各个子问题，其中如果T[i]==S[j]，那么dp[i][j]=dp[i][j-1] + dp[i-1][j-1]；否则，dp[i][j]=dp[i][j-1]。其中第一行跟第一列都可以实现初始化。代码：

class Solution {
public:
int numDistinct(string S, string T) {
if (S.empty() || T.empty()){
return 0;
}
int len_S = S.size();
int len_T = T.size();
vector<int> temp(len_S);
vector<vector<int> > dp(len_T,temp);
if(T[0] == S[0])    dp[0][0] = 1;
else    dp[0][0] = 0;
for(int i = 1; i < len_S; ++i){
if(T[0] == S[i]){
dp[0][i] = dp[0][i-1] + 1;
}else{
dp[0][i] = dp[0][i-1];
}
}
for(int i = 1; i < len_T; ++i){
dp[i][0] = 0;
}
for(int i = 1; i < len_T; ++i){
for(int j = 1; j < len_S; ++j){
if(T[i] != S[j]){
dp[i][j] = dp[i][j-1];
}else{
dp[i][j] = dp[i][j-1] + dp[i-1][j-1];
}
}
}
return dp[len_T-1][len_S-1];
}
};