Given a string s, return the number of distinct non-empty subsequences of s. Since the answer may be very large, return it modulo 1e9 + 7.
A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not.
Example 1
Input: s = "abc"
Output: 7
Explanation: The 7 distinct subsequences are "a", "b", "c", "ab", "ac", "bc", and "abc".
Example 2
Input: s = "aba"
Output: 6
Explanation: The 6 distinct subsequences are "a", "b", "ab", "aa", "ba", and "aba".
Example 3
Input: s = "aaa"
Output: 3
Explanation: The 3 distinct subsequences are "a", "aa" and "aaa".
Constraints
- 1 <= s.length <= 2000
- s consists of lowercase English letters.
Solution
f[i] 表示第 i 个字符作为最后一个字符时的子序列个数，last[i] 是随着遍历一起更新的而不是一开始就确定好的。
初始化 f[i] = 1 ，每次都加一遍 last[i] 然后更新 last[i]
是前面的全都生成好了，然后加上下一个字符以后的情况。
就是当前 f[i] 是前面已经遍历过的子序列之和，但是是用 f[last[j]] 来加和。
#define MOD 1000000007
int distinctSubseqII(char * s){
int i, j, n, ans, last[26];
n = strlen(s);
int f[n];
for (i = 0; i < 26; i++) last[i] = -1;
for (i = 0; i < n; i++) f[i] = 1;
for (i = 0; i < n; i++) {
for (j = 0; j < 26; j++) {
if (last[j] != -1) f[i] = (f[i] + f[last[j]]) % MOD;
}
last[s[i] - 'a'] = i;
}
ans = 0;
for (i = 0; i < 26; i++) {
if (last[i] != -1) ans = (ans + f[last[i]]) % MOD;
}
return ans;
}
