将最长回文子序列的图片文件夹改为英文
labuladong
@ -73,7 +73,7 @@ Now let's talk about the Longest Palindrome Subsequence (LPS) problem to explain
|
||||
|
||||
We have solve the "Longest Palindrome Substring" problem before. This time, the difficulty is increased by finding the length of the Longest Palindrome Subsequence instead of substring:
|
||||
|
||||
|
||||
|
||||
|
||||
In this question, **we define `dp[i][j]` as the length of the longest palindrome subsequence within the substring `s[i..j]`**. Please remember this definition so as to understand the algorithm.
|
||||
|
||||
@ -83,7 +83,7 @@ Specifically, if we want to find `dp[i][j]`, suppose you have already got the re
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
The answer is yes! It depends on the characters of `s[i]` and `s[j]`:
|
||||
|
||||
@ -91,11 +91,11 @@ The answer is yes! It depends on the characters of `s[i]` and `s[j]`:
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
**If they are not equal**, it means that they **cannot appear at the same time** in the longest palindrome subsequence of `s[i..j]`. Therefore, we add them **separately** to `s[i+1..j-1] ` to see which substring produces a longer palindrome subsequence:
|
||||
|
||||
|
||||
|
||||
|
||||
The code of the above two cases can be written like this:
|
||||
|
||||
@ -118,9 +118,9 @@ Since `i`must be less than or equal to `j`, for those locations where `i > j`, t
|
||||
|
||||
In addition, look at the state transition equation we just got. To find `dp[i][j]`, you need to know `dp[i+1][j-1]`, `dp[i+1][j]` and`dp[i][j -1]` these three values. And look at the base case we determined, this is how the DP array looks like after being filled:
|
||||
|
||||
|
||||
|
||||
|
||||
**In order to guarantee that before each calculation of `dp[i][j]`, the values in the left, down and right direction have been calculated, we can only traverse it diagonally or reversely**:
|
||||
**In order to guarantee that before each calculation of `dp[i][j]`, the values in the left, down and right direction have been calculated, we can only traverse it diagonally or reversely**:
|
||||
|
||||
Here I choose to traverse reversely. The code is as follows:
|
||||
|
||||
|
||||
|
Before Width: | Height: | Size: 147 KiB After Width: | Height: | Size: 147 KiB |
|
Before Width: | Height: | Size: 145 KiB After Width: | Height: | Size: 145 KiB |
|
Before Width: | Height: | Size: 152 KiB After Width: | Height: | Size: 152 KiB |
|
Before Width: | Height: | Size: 157 KiB After Width: | Height: | Size: 157 KiB |
|
Before Width: | Height: | Size: 184 KiB After Width: | Height: | Size: 184 KiB |