Leetcode所有题目和解答（多语言实现）.rar

## Guessing Lets start with `rabbbit` and `rabbit`. To make the problem easier, start from one char, `distinct subsequences` of `rabbbit` and `r`. `distinct subsequences` below, by eyeball: * `r` and `r` has 1 `distinct subsequences` * `a` and `r` has 0 `distinct subsequences` These two are easy to be understood. * `ra` and `r` has 1 `distinct subsequences` * `rab` and `r` has 1 `distinct subsequences` * `rabbbbbbbbbbbb` and `r` has 1 `distinct subsequences` These three show that `r` plus any junk chars will not change `distinct subsequences` * `rr` and `r` has 2 `distinct subsequences` * `rar` and `r` has 2 `distinct subsequences` * `rara` and `r` has 2 `distinct subsequences` * `rarar` and `r` has 3 `distinct subsequences` `rar` and `r` is same as `rr` and `r`, because `a` there has nothing to do with the number of `distinct subsequences`, it is a junk like above. In another word, `rar` and `r` can be converted to `ra` + `r` and `r`, that is `1 + 1`. ### Put them into table the value `P[i][j]` means that `S[0..i]` and `T[0..j]` has `P[i][j]` `distinct subsequences`. the table with only one char ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ ``` Enlarge the table ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | a | | | | | | | | +---+---+---+---+---+---+---+---+ | b | | | | | | | | +---+---+---+---+---+---+---+---+ | b | | | | | | | | +---+---+---+---+---+---+---+---+ | i | | | | | | | | +---+---+---+---+---+---+---+---+ | t | | | | | | | | +---+---+---+---+---+---+---+---+ ``` Some grids here are easy to be filled, the grids which `j > i`. That is, T is longer than S, there will be no `distinct subsequences`. ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | a | 0 | | | | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | | | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 0 | | | | | +---+---+---+---+---+---+---+---+ | i | 0 | 0 | 0 | 0 | | | | +---+---+---+---+---+---+---+---+ | t | 0 | 0 | 0 | 0 | 0 | | | +---+---+---+---+---+---+---+---+ ``` `j == i` grids are easy too, the S and T must be the same if it wants to have 1 `distinct subsequences`. ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | a | 0 | 1 | | | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 1 | | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 0 | 1 | | | | +---+---+---+---+---+---+---+---+ | i | 0 | 0 | 0 | 0 | 0 | | | +---+---+---+---+---+---+---+---+ | t | 0 | 0 | 0 | 0 | 0 | 0 | | +---+---+---+---+---+---+---+---+ ``` This pictrue is just encouraging you, as there is only a few grids left to be filled. ### `rab` and `ra` This problem looks familiar. It is the `ra` and `r`. as `b` is a junk char, `rab` should have the same number as `ra` and `ra`. Then the table is like ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | a | 0 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 1 | | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 0 | 1 | | | | +---+---+---+---+---+---+---+---+ | i | 0 | 0 | 0 | 0 | 0 | | | +---+---+---+---+---+---+---+---+ | t | 0 | 0 | 0 | 0 | 0 | 0 | | +---+---+---+---+---+---+---+---+ ``` ### `rabb` and `rab` This time, add a new `b` to both `rab` and `ra`. #### Use the new `b` in subsequences This time they have same letter, `b` , at the end. It is not a junk. This time, remove `b` from both S and T, then they become `rab` and `ra`. that is, this will take `rab` and `ra` `distinct subsequences`. #### Dont use the new `b` Treat `b` as junk, it has `rab` and `rab` `distinct subsequences`. #### Sum up two choices Use or not will result two set of `distinct subsequences`, so sum them up. `rabb` and `rab` = `rab` and `ra` + `rab` and `rab`. ``` +---+---+---+---+---+---+---+---+ | | r | a | b | b | b | i | t | +---+---+---+---+---+---+---+---+ | r | 1 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | a | 0 | 1 | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 1 | 2 | | | | +---+---+---+---+---+---+---+---+ | b | 0 | 0 | 0 | 1 | | | | +---+---+---+---+---+---+---+---+ | i | 0 | 0 | 0 | 0 | 0 | | | +---+---+---+---+---+---+---+---+ | t | 0 | 0 | 0 | 0 | 0 | 0 | | +---+---+---+---+---+---+---+---+ ``` ## General form the grid `P[i][j]` has 1 choice when `S[i] != T[j]`, this only adds some junk chars. ``` P[i][j] = P[i - 1][j] ``` the grid `P[i][j]` has 2 choice when `S[i] == T[j]` ``` P[i][j] = P[i - 1][j] // adds as junk chars. + P[i - 1][j - 1] // use the new char in the new subsequence ``` Fill up the table and the bottom right grid is the answer.