Distinct Subsequences
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).
Example 1:
Input: S = “rabbbit”, T = “rabbit”
Output: 3
Explanation:
As shown below, there are 3 ways you can generate “rabbit” from S.
(The caret symbol ^ means the chosen letters)
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
Example 2:
Input: S = “babgbag”, T = “bag”
Output: 5
Explanation:
As shown below, there are 5 ways you can generate “bag” from S.
(The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^
解析
找到字符串s中子串t的個數。
自己的解法
設一個二維數組dp[t.len][s.len]，dp[i][j]表示字符串s[0,j]有子串t[0,i]的個數，只有s[j]==t[i]時，纔會>0；
當匹配到t的第i+1個字符時，當前與t[i+1]相同的字符位置s[j]的dp[i+1][j]是上一個匹配字符的個數之和，即dp[i+1][j] = dp[i][k] k<j；
最後結果爲t的最後一個字符全部匹配的結果之和。
class Solution {
public:
int numDistinct(string s, string t) {
int len1 = s.size(),len2 = t.size();
if(len1 == 0 || len1 < len2)
return 0;
vector<vector<long> > dp(len2, vector<long>(len1,0));
for(int i=0;i<len2;i++){
for(int j=0;j<len1;j++){
if(t[i] == s[j]){
if(i == 0)
dp[0][j] = 1;
else{
for(int k=0;k<j;k++)
dp[i][j] += dp[i-1][k];
}
}
}
}
int res = 0;
for(int j=0;j<len1;j++)
res += dp[len2-1][j];
return res;
}
};
動態規劃
同樣設置一個動態規劃數組，dp[i][j] 表示s中範圍是 [0, i] 的子串中能組成t中範圍是 [0, j] 的子串的子序列的個數。
邊界條件：子序列爲空時，返回1，即dp[0][j] = 1
dp[i][j] 時，dp[i][j] >= dp[i][j - 1] 總是成立，
當 T[i - 1] == S[j - 1] 時，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，
若不等， dp[i][j] = dp[i][j - 1]，
綜合以上，遞推式爲：
dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)
class Solution {
public:
int numDistinct(string s, string t) {
int len1 = s.size(),len2 = t.size();
if(len1 == 0 || len1 < len2)
return 0;
vector<vector<long> > dp(len2+1, vector<long>(len1+1,0));
for(int i=0;i<=len2;i++){
for(int j=i;j<=len1;j++){
if(i == 0) dp[i][j] = 1;
else{
dp[i][j] = t[i-1] == s[j-1] ? dp[i][j-1]+dp[i-1][j-1] : dp[i][j-1];
}
}
}
return dp[len2][len1];
}
};