[LeetCode] Longest Palindromic Substring-优快云博客

本文探讨了回文问题的解决策略，最初使用递归算法导致效率低下，通过将原始字符串反转并查找最大公共子串的方法进行优化，虽然存在代码重复，但实现了性能提升。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

回文问题，一开始搞了个递归算法，重复计算太多，结果超时了。

目前改成先把原来的字符串反转，然后查找两个字符串最大公共字串的问题，勉强通过吧，代码也有重复，有点懒了...

class Solution {
public:
    string longestPalindrome(string s) {
        int length = s.size();
        char *s1 = new char[length];
        char *s2 = new char[length];
        for (int i = 0; i < length; ++i) {
            s1[i] = s[i];
            s2[length - 1 - i] = s[i];
        }
        string sr(s2, length);

        int maxi = 0;
        int start = 0;
        int num = 0;
        bool mark = false;
        string result = "";
        for (int j = 1; j <= length; ++j) {

            for (int m = 0; m < j; ++m) {
                if (s1[m] == s2[length - j + m]) {
                    if (mark == false) {
                        mark = true;
                        start = m;
                    }
                    num++;
                }
                else {
                    if (mark == true) {
                        mark = false;
                        if (num > maxi) {
                            maxi = num;
                            result = s.substr(start, num);
                        }
                    }
                    num = 0;
                }
            }
            if (num > maxi) {
                maxi = num;
                result = s.substr(start, num);
            }
            num = 0;
            mark = false;
        }
        num = 0;
        mark = false;
        for (int j = 1; j <= length; ++j) {

            for (int m = 0; m < j; ++m) {
                if (s2[m] == s1[length - j + m]) {
                    if (mark == false) {
                        mark = true;
                        start = m;
                    }
                    num++;
                }
                else {
                    if (mark == true) {
                        mark = false;
                        if (num > maxi) {
                            maxi = num;
                            result = sr.substr(start, num);
                        }
                    }
                    num = 0;
                }
            }
            if (num > maxi) {
                maxi = num;
                result = sr.substr(start, num);
            }
            num = 0;
            mark = false;
        }
        return result;
    }
};