原题链接在这里：https://leetcode.com/problems/distinct-subsequences/
题目：
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: 3Explanation: 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: 5Explanation: 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^^^
题解：
Let dp[i,j]表示s[0....i]有多少种方法可以减成t[0.......j].
If s.charAt(i-1) == t.charAt(j-1),
dp[i][j] = dp[i-1][j-1] (用上最新的match, 之前s[0....i-1]减到t[0...j-1]的方法都用最后一个match, 所以总共的方法数不变) + dp[i-1][j] (不用当前match, 之前s[0...i-1]减到t[0...j]的方法即可).
若s.charAt(i-1) != t.charAt(j-1),
不match的话只能是之前s[0...i-1]减到t[0...j]的方法.
Time Complexity: O(m*n).m = s.length().n = t.length().
Space:O(m*n).
AC Java:
1 class Solution { 2 public int numDistinct(String s, String t) { 3 int m = s.length(); 4 int n = t.length(); 5 int [][] dp = new int[m+1][n+1]; 6 for(int i = 0; i<=m; i++){ 7 dp[i][0] = 1; 8 } 9 10 for(int i = 1; i<=m; i++){ 11 for(int j = 1; j<=n; j++){ 12 if(s.charAt(i-1) == t.charAt(j-1)){ 13 dp[i][j] = dp[i-1][j-1] + dp[i-1][j]; 14 }else{ 15 dp[i][j] = dp[i-1][j]; 16 } 17 } 18 } 19 20 return dp[m][n]; 21 } 22 }