Question

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
    ^^^

Thinking:

  • Method: dp
    • 当t长度为0时， 无论i长度为什么均为1,。
    • 当s[i] ！= t[j]时，当前结果为前面出现的次数。dp[i][j] = dp[i - 1][j]。
    • 当s[i] == t[j]时，结果为t长度不变dp[i - 1][j]，前面成立的次数加上dp[i - 1][j - 1]。
class Solution {
    public int numDistinct(String s, String t) {
        int sLen = s.length();
        int tLen = t.length();
        int[][] dp = new int[sLen + 1][tLen + 1];
        for(int i = 0; i < sLen; i++)
            dp[i][0] = 1;
        for(int i = 1; i <= sLen; i++){
            for(int j = 1; j <= tLen; j++){
                dp[i][j] += dp[i - 1][j];
                if(s.charAt(i - 1) == t.charAt(j - 1)){
                    dp[i][j ] = dp[i - 1][j] + dp[i - 1][j - 1];
                }
            }
        }
        return dp[sLen][tLen];
    }
}
  • 延伸：s,t两个字符串，能否从s中移除一些字符变成b。 
    public static boolean stringChange(String s, String t){
      int sLen = s.length();
      int tLen = t.length();
      boolean[][] dp = new boolean[sLen + 1][tLen + 1];
      for(int i = 0; i <= sLen; i++)
          dp[i][0] = true;
      for(int i = 1; i <= sLen; i++){
          for(int j = 1; j <= tLen; j++){
              if(s.charAt(i - 1) != t.charAt(j - 1)){
                  dp[i][j] |= dp[i - 1][j];	//在t不变的情况下，如果前面成功了就一直成功。
              }else{
                  dp[i][j] = dp[i - 1][j] | dp[i - 1][j - 1];//如果相同仍然是前面成功就一直成功，但是要多加一个条件，就是S，T长度各减一的情况下如果成功，当前也是成功的，所以同样成功。
              }
          }
      }
      for(boolean[] bs : dp){
          for(boolean b : bs)
              System.out.print(b + "   ");
          System.out.println();
      }
      return dp[sLen][tLen];
  }

二刷

class Solution {
    public int numDistinct(String s, String t) {
        if(s == null || t == null) return 0;
        int sLen = s.length(), tLen = t.length();
        int[][] dp = new int[sLen + 1][tLen + 1];
        dp[0][0] = 1;
        for(int i = 1; i <= sLen; i++)
            dp[i][0] = 1;
        char[] sArr = s.toCharArray();
        char[] tArr = t.toCharArray();
        for(int i = 1; i <= sLen; i++){
            char c1 = sArr[i - 1];
            for(int j = 1; j <= tLen; j++){
                char c2 = tArr[j - 1];
                dp[i][j] = dp[i - 1][j];
                if(c1 == c2) dp[i][j] += dp[i - 1][j - 1];
            }
        }
        return dp[sLen][tLen];
    }
}