LeetCode 940. Distinct Subsequences II

Given a string S, count the number of distinct, non-empty subsequences of S .
Since the result may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: "abc"
Output: 7
Explanation: The 7 distinct subsequences are "a", "b", "c", "ab", "ac", "bc", and "abc".
 

Example 2:

Input: "aba"
Output: 6
Explanation: The 6 distinct subsequences are "a", "b", "ab", "ba", "aa" and "aba".

Example 3:

Input: "aaa"
Output: 3
Explanation: The 3 distinct subsequences are "a", "aa" and "aaa".

Note:
S contains only lowercase letters.
1 <= S.length <= 2000

假設dp[i]存的是當前答案的數字

    public int distinctSubseqII(String s) {
        int m = (int)Math.pow(10,9) + 7;
        List<List<Integer>> map = new ArrayList<>();
        for (int i =0; i< 26; i++){
            map.add(new ArrayList<>());
        }
        int [] dp = new int[s.length()];
        for (int i = 0; i < s.length(); i++) {
            if (i == 0) {
                dp[i] = 1;
            } else {
                dp[i] = 2 * dp[i-1] + 1;
                if (map.get(s.charAt(i) - 'a').size() != 0) {
                    for (int index : map.get(s.charAt(i) - 'a')) {
                        if (index > 0)  {
                            dp[i] = dp[i] - dp[index] + dp[index - 1];
                        } else {
                            dp[i] = dp[i] - dp[index];
                        }
                    }
                }
            }
            
            //System.out.println("dp["+i+"]: " + dp[i]);
            map.get(s.charAt(i) - 'a').add(i);
        }
        return dp[s.length() - 1];
    }
 

a, b, a, b
a b aa bb
ab ba aab
aba bab
abab
abb

假設dp[i]存的是當前新增的數目

    public int distinctSubseqII(String s) {
        int m = (int)Math.pow(10,9) + 7;
        int[] dp = new int[s.length()];
        for(int i = 0; i < s.length(); ++i) {
            dp[i] = 1;
            for(int j = 0; j < i; ++j) {
                if(s.charAt(i) != s.charAt(j)) {
                    dp[i] += dp[j];    
                    dp[i] %= m;
                }                
            }
        }
        int sum = 0;
        for(int i = 0;i < dp.length; ++i) {
            sum += dp[i];
            sum %= m;
        }
        return sum;
    }