Topic links:
https://leetcode-cn.com/problems/distinct-subsequences/submissions/
Ideas:
A title dp
State transition equation is:
if (s [j] == t [i]) dp [i] [j] = dp [i-1] [j-1] = dp [i] [j-1] // meaning: If the string the same as the corresponding character, transfer status, the number of sub-sets of strings before this appeared in the substring is the number of times a substring that appear on the plus before the number of occurrences of the string (do not know did not make it clear, I understand / thinking)
else dp [i] [j] = dp [i] [j-1], if not equal, in this cell must be the number of the current substring occurring in the set, i.e. on a cell's value (because no new same string)
Forms illustrated as follows:
(Empty string certainly appear once in all sub-string, and the first column because the original string is empty, there is no corresponding appearance - bad language, loud noise may not be clear, but because of the plug-understand / Behind his. )
| t \ s | '' | b | a | b | g | b | a | g |
| '' | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| b | 0 | 1 | 1 | 2 | 2 | 3 | 3 | 3 |
| a | 0 | 0 | 1 | 1 | 1 | 1 | 4 | 4 |
| g | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 5 |
Here is the source code:
package leetcode; public class NumDistinct { /** * dynamic programming * @param s * @param t * status transfer formula is : * if s==t dp[i][j] = dp[i-1][j-1] + dp[i][j-1]:means the sum of combinations and already exist * if s!=t dp[i][j] = dp[i][j-1] * @return */ public int numDistinct(String s, String t) { if(s.length()==0&&t.length()==0)return 0; int[][] dp = new int[t.length()+1][s.length()+1];//build status transfer array for(int i = 0;i<dp.length;++i){ dp[i][0] = 0; } for(int j = 0;j<dp.length;++j){ dp[0][j] = 1; }//initialize char[] s_temp = s.toCharArray(); char[] t_temp = t.toCharArray(); for(int i = 1;i<dp.length;++i){ for(int j = 1;j<dp[0].length;++j){ if(t_temp[i-1]==s_temp[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i][j-1]; else dp[i][j] = dp[i][j-1]; } } return dp[s.length()][t.length()]; } }
Code has been ac
I hope to be helpful
the above