给定字符串S和T,S通过删除某些位置的字符得到T的话,就记作一种subSequence。返回总共有几种。
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
.
思路一:分分钟就写出了递归的算法。
class Solution { public: int numDistinct(string S, string T) { if (S.size() < T.size()) return 0; if (S.size() == T.size() && S == T) return 1; // 都为空的时候返回什么呢??? if (S[0] != T[0]) return numDistinct(S.substr(1), T); else return numDistinct(S.substr(1), T.substr(1)) + numDistinct(S.substr(1), T); } };
Time Limited
思路二:这样的题都是可以用动态规划解决的。
用dp[i][j]记录S的前i个和T的前j个的符合个数,那么最后目标就是dp[S.size()][T.size()];
初始化,j = 0 时候,dp[i][0] = 1,因为所有的都可以通过删除所有变成空字符,并且只有一种。
递推式子如下了:
i和j都从1开始,且j不能大于i,因为匹配的长度至少为1开始,j大于i无意义
如果 i == j 那么 dp[i][j] = S.substr(0, i) == T.substr(0, j);
如果 i != j 分两种情况
S[i-1] != T[j-1] 时,也就是加入不加入i的影响是一样的,那么 dp[i][j] = dp[i - 1][j];
S[i-1] == T[j-1] 时,那么当前字符可选择匹配或者是不匹配,所以dp[i][j] = dp[i - 1][j -1] + dp[i - 1][j];
整理成代码:
class Solution { public: int numDistinct(string S, string T) { int lenS = S.size(), lenT = T.size(); if (lenS < lenT) return 0; if (lenT == 0) return lenS; int dp[lenS + 1][lenT + 1]; //memset(dp, 0, sizeof(dp)); for (int i = 0; i <= lenS; ++i) dp[i][0] = 1; for (int i = 1; i <= lenS; ++i) for (int j = 1; j <= lenT && j <= i; ++j) { if (i == j) dp[i][j] = S.substr(0, i) == T.substr(0, j); else if (S[i - 1] != T[j - 1]) dp[i][j] = dp[i - 1][j]; else dp[i][j] = dp[i - 1][j -1] + dp[i - 1][j]; } return dp[lenS][lenT]; } };
如果,初始化用了memset了,那么if(i==j)语句就可以省略了