392. Is Subsequence

Given two strings s and t, return true if s is a subsequence of t , or false otherwise.

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).

代码：
1.DP

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

115. Distinct Subsequences

Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that the answer fits on a 32-bit signed integer.

题目解析：

• dp[i][j]：以i-1为结尾的s子序列中出现以j-1为结尾的t的个数为dp[i][j]。
• 当s[i - 1] 与 t[j - 1]相等时，dp[i][j]可以有两部分组成。
  • 一部分是用s[i - 1]来匹配，那么个数为dp[i - 1][j - 1]。即不需要考虑当前s子串和t子串的最后一位字母，所以只需要 dp[i-1][j-1]。
  • 一部分是不用s[i - 1]来匹配，个数为dp[i - 1][j]。
• 当s[i - 1] 与 t[j - 1]不相等时，dp[i][j]只有一部分组成，不用s[i - 1]来匹配（就是模拟在s中删除这个元素），即：dp[i - 1][j]

代码：

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 j = 1; j < t.length(); j++) {
            dp[0][j] = 0;
        }
        for (int i = 0; i < s.length(); i++) {
            for (int j = 0; j < t.length(); j++) {
                if (s.charAt(i) == t.charAt(j)) {
                    dp[i+1][j+1] = dp[i][j] + dp[i][j+1];
                } else {
                    dp[i+1][j+1] = dp[i][j+1];
                }
            }
        }
        return dp[s.length()][t.length()];
    }
}