●  392.判断子序列 Is Subsequence - LeetCode

dp[i][j] 以i - 1为结尾的和以 j - 1为结尾的子串的相同子序列长度

if (s[i - 1] == t[j - 1]）dp[i][j] = dp[i - 1][j - 1] + 1

else dp[i][j] = dp[i][j - 1]  

for (int i = 1; i <= s.length(); i++)

        for (int j = 1; j <= t.length(); j++) 

dp[s.length()[t.length()] = s.length();

class Solution {public boolean isSubsequence(String s, String t) {int[][] dp = new int[s.length() + 1][t.length() + 1];dp[0][0] = 0;for (int i = 1; i <= s.length(); i++) {for (int j = 1; j <= t.length(); j++) {if (s.charAt(i - 1) == t.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = dp[i][j - 1];}}}if (dp[s.length()][t.length()] == s.length()){return true;} else {return false;}}
}

                

●  115.不同的子序列  Distinct Subsequences - LeetCode

dp[i][j] 以i-1为结尾的s中有以j-1为结尾的t的个数为

if (s[i - 1] == t[j - 1])dp[i][j] = dp[i - 1][j - 1] + 1 + dp[i - 1][j];

else dp[i][j] = dp[i - 1][j];

dp[i][0] = 1;

dp[0][j] = 0;

dp[0][0] = 0;

for (int i = 1; i <= s.length(); i++)

        for (int j = 1; j <= t.length(); j++)

class Solution {public int numDistinct(String s, String t) {int[][] dp = new int[s.length() + 1][t.length() + 1];for (int i = 0; i < s.length(); i++) {dp[i][0] = 1;}for (int i = 1; i < s.length() + 1; i++) {for (int j = 1; j < t.length() + 1; j++) {if (s.charAt(i - 1) == t.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];} else {dp[i][j] = dp[i - 1][j];}}}return dp[s.length()][t.length()];}
}