Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.
Example 1:
Input:
"bbbab"
Output:
4
One possible longest palindromic subsequence is "bbbb".
第一感觉就是DP，可惜写不出来，只想到用一维DP的思想，前i位的LPS，但总觉得这么想没什么意义，因为跟结尾也有关啊。
看了答案：
dp[i][j]: the longest palindromic subsequence's length of substring(i, j)
State transition:
dp[i][j] = dp[i+1][j-1] + 2 if s.charAt(i) == s.charAt(j)
otherwise, dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1])
Initialization: dp[i][i] = 1
画个图比较清楚
public class Solution {
public int longestPalindromeSubseq(String s) {
int[][] dp = new int[s.length()][s.length()];
for (int i = s.length() - 1; i >= 0; i--) {
dp[i][i] = 1;
for (int j = i+1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = dp[i+1][j-1] + 2;
} else {
dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
}
}
}
return dp[0][s.length()-1];
}
}
从后往前循环是因为dp[i][j]代表从i到j的LPS。如果从前往后，就要改成dp[j][i]。
