最长回文子串的动态规划写法

最长回文子串O(n^2)的动态规划方程

用f[x][y] 表示从字符串下标x到下标y的回文长度

那么如果要求连续的回文子串则如果s[x] != s[y]  f[x][y] = 0;  如果相等则 如果f[x+1][y-1] > 0  则f[x][y] = f[x+1][y-1] + 2

由于这里出现了x+1跟y-1的情况，所以x,y的距离至少为1

int f[2002][2002];
class Solution {
public:
    int longestPalindrome(string word1, string word2) {
        string words = word1 + word2;
        
        memset(f, 0, sizeof(f));
        int len = words.length();
        for (int i = 0; i <= len; ++i)
            f[i][i] = 1;

        int maxVal = 0;
        for (int i = 1; i < len; ++i) {
            for (int j = 0; j < len - i; ++j) {if (words[j] == words[j + i] && f[j + 1][j + i - 1] > 0) {
                    f[j][j + i] = f[j + 1][j + i - 1] + 2;
                    maxVal = max(maxVal, f[j][j+i]);
                }
            }
        }

        return maxVal;
    }
};

 

 

那么如果不是求连续的回文子串则给代码增加对应的判断即可

int f[2002][2002];
class Solution {
public:
    int longestPalindrome(string word1, string word2) {
        string words = word1 + word2;
        
        memset(f, 0, sizeof(f));
        int len = words.length();
        for (int i = 0; i <= len; ++i)
            f[i][i] = 1;

        int maxVal = 0;
        for (int i = 1; i < len; ++i) {
            for (int j = 0; j < len - i; ++j) {
                f[j][j+i] = max(f[j+1][j+i], f[j][j+i-1]);
                if (words[j] == words[j + i]) {
                    f[j][j + i] = f[j + 1][j + i - 1] + 2;
                    maxVal = max(maxVal, f[j][j+i]);
                }
            }
        }

        return maxVal;
    }
};

 

leetcode的5688即求最长回文子串可用如上的解法

int f[2002][2002];
class Solution {
public:
    int longestPalindrome(string word1, string word2) {
        string words = word1 + word2;
        
        memset(f, 0, sizeof(f));
        int len = words.length();
        for (int i = 0; i <= len; ++i)
            f[i][i] = 1;

        int maxVal = 0;
        for (int i = 1; i < len; ++i) {
            for (int j = 0; j < len - i; ++j) {
                f[j][j+i] = max(f[j+1][j+i], f[j][j+i-1]);
                if (words[j] == words[j + i]) {
                    f[j][j + i] = f[j + 1][j + i - 1] + 2;
                    if (j < word1.length() && j + i >= word1.length())
                        maxVal = max(maxVal, f[j][j+i]);
                }
            }
        }

        return maxVal;
    }
};