## 题目描述：

LeetCode 516. Longest Palindromic Subsequence

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".

Example 2:
Input:

```"cbbd"
```

Output:

```2
```

One possible longest palindromic subsequence is "bb".

## 解题思路：

```dp[i][j] = dp[i + 1][j - 1] + 2           if s[i] == s[j]
dp[i][j] = max(dp[i][j - 1], dp[i + 1][j])    otherwise```

## Java代码：

``````public class Solution {
public int longestPalindromeSubseq(String s) {
int size = s.length();
int[][] dp = new int[size][size];
for (int i = size - 1; i >= 0; i--) {
dp[i][i] = 1;
for (int j = i + 1; j < size; 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][size - 1];
}
}
``````

```令s' = reversed(s), size = len(s)

dp[i][j]表示s[0 .. i]与s'[0 .. j]的最长公共子序列的长度

## Java代码：

``````public class Solution {
public int longestPalindromeSubseq(String s) {
int size = s.length();
int[][] dp = new int[size + 1][size + 1];
for (int i = 1; i <= size; i++) {
for (int j = 1; j <= size; j++) {
if (s.charAt(i - 1) == s.charAt(size - j)) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);
}
}
}
int ans = s.length() > 0 ? 1 : 0;
for (int m = 0; m < size; m++) {
ans = Math.max(dp[m][size - m] * 2, ans);
if (m > 0) ans = Math.max(dp[m - 1][size - m] * 2 + 1, ans);
}
return ans;
}
}
``````

Pingbacks已关闭。