LintCode 667: Longest Palindromic Subsequence (DP 经典题)

经典DP题。我们最后需要返回dp[0][len-1]，所以两个for loop是下面的代码：

        for (int i = len - 1; i >= 0; --i) {
            cout<<"i="<<i<<endl;
            for (int j = i + 1; j < len; ++j) {

对于输入 “bbbab”, dp[][] 值为:

1 2 3 3 4
0 1 2 2 3
0 0 1 1 3
0 0 0 1 1
0 0 0 0 1

注意下面两种写法不对：
1)
for (int i = 0; i < len - 1; ++i) {
for (int j = i + 1; j < len; ++j) {

It is wrong as dp[0][len - 1] should be the final result.

2. for (int i = 0; i < len - 1; ++i) {
   for (int j = 0; j < len; ++j) {
   It is wrong as dp[i+1][j] will exceed the boundary.

代码如下：

class Solution {
public:
    /**
     * @param s: the maximum length of s is 1000
     * @return: the longest palindromic subsequence's length
     */
    int longestPalindromeSubseq(string &s) {
        int len = s.size();
        if (len <= 1) return len;
        
        vector<vector<int>> dp(len, vector<int>(len, 0));
        
        for (int i = 0; i < len; ++i) {
            dp[i][i] = 1;
        }
        for (int i = len - 1; i >= 0; --i) {
            for (int j = i + 1; j < len; ++j) {
                if (s[i] == s[j]) {
                    dp[i][j] = dp[i + 1][j - 1] + 2;
                } else {
                    dp[i][j] = max(dp[i][j - 1], dp[i + 1][j]);
                }
            }
        }
        return dp[0][len - 1];
    }
};