题目:给两个字符串S和T,判断T在S中出现的次数。
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1:递归(TLE)
如果当前字符相同,结果加上S和T在该index之后的匹配方法数
如果当前字符不同,将S的指针向后移,递归计算
class Solution { private: int cnt; int len_s; int len_t; public: Solution():cnt(0){} void Count(string S,string T, int idx_ss, int idx_ts){ if(idx_ts == len_t){ cnt++; return; } int i,j,k; for (i=idx_ss; i<len_s; i++) { if (S[i] == T[idx_ts]) { Count(S, T, i + 1, idx_ts + 1); } } } int numDistinct(string S, string T) { len_s = S.length(); len_t = T.length(); Count(S, T, 0, 0); return cnt; } };
-----------------------------------------------------
思路2:DP
如果当前字符相同,dp[i][j]结果等于用S[i](dp[i-1][j-1])和不用S[i](dp[i-1][j])方法数求和
如果当前字符不同,dp[i][j] = dp[i-1][j]
class Solution { private: int len_s; int len_t; public: int Count(string S,string T){ int i,j; int dp[len_s][len_t]; memset(dp, 0, sizeof(dp)); if (S[0]==T[0]) { dp[0][0] = 1; } for(i=1;i<len_s;i++){ dp[i][0] = dp[i-1][0]; if (T[0]==S[i]) { dp[i][0]++; } } for (i=1; i<len_s; i++) { for (j=1; j<len_t && j<=i; j++) { if (S[i]!=T[j]) { dp[i][j] = dp[i-1][j]; //cout<<dp[i-1][j]<<endl; } else{ dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; //dp[i-1][j-1]: use S[i], as S[i]==T[j] //dp[i-1][j] : don‘t use S[i] //cout<<dp[i][j]<<endl; } } } return dp[len_s-1][len_t-1]; } int numDistinct(string S, string T) { len_s = S.length(); len_t = T.length(); return Count(S, T); } };
时间: 2024-10-28 19:14:34