力扣 题目5- 最长回文子串(100道完结)

题目

题解

1.暴力解法 从前往后遍历途中对 以i为中心对称遍历 和 i也有对称数的对称遍历

2.动态规划 一个回文子串 意味着将两端去掉依然是回文子串 所以我们使用两层vector 记录从开始位置到结束位置是否是回文字符

当s[j]==s[i]时 就去看res[i + 1][j - 1] 是否也为true 是则res[i][j] = true;

单个字符或者两个时直接为true   j - i <= 1->res[i][j] = true;

代码

1.

#include<iostream>
#include<string>
using namespace std;
class Solution {
public:
    string longestPalindrome(string s) {
        string result="";
        string tem = "";
        int left = 0;
        int right = 0;
        for (int i = 0; i < s.size(); i++) {
        //1.中心对称
            tem = s[i];
            left = i-1;
            right = i + 1;
            while (left>-1&& right<s.size()&&s[left]==s[right])
            {
                tem = s[left]+tem+s[right];
                left--;
                right++;
            }
            if (result.size() < tem.size()) {
                result = tem;
            }
            if (tem.size()== s.size()) {
                return s;
            }

            tem = "";
        //2.双向对称
            left = i;
            right = i + 1;
            while (left > -1 && right < s.size() && s[left] == s[right])
            {
                tem = s[left] + tem + s[right];
                left--;
                right++;
            }
            if (result.size() < tem.size()) {
                result = tem;
            }
            if (tem.size() == s.size()) {
                return s;
            }
        }
        return result;
    }
};

int main() {
    Solution sol;
    string s = "cbbd";
    string reslut=sol.longestPalindrome(s);
    cout << reslut << endl;
}

2.

#include<iostream>
#include<vector>
#include<string>
using namespace std;
//vector<vector<string>>(s.size()) = { vector<string>(s.size(),"233") }
class Solution {
public:
    string longestPalindrome(string s) {
        int maxlenth = 0;
        int left = 0;
        int right = 0;
        //vector<vector<vector<string>>> res(s.size(), vector<vector<string>>(s.size()) );
        vector<vector<int>> res(s.size(), vector<int>(s.size(),0));
        for (int i = s.size() - 1; i >= 0; i--) {
            for (int j = i; j < s.size(); j++) {
                if (s[j]==s[i]) {
                    //j与i相等 即同一个字符 符合回文
                    if (j - i <= 1) { 
                        res[i][j] = true;
                    }
                    else if (res[i + 1][j - 1]) { 
                        res[i][j] = true;
                    }
                }
                if (res[i][j] && j - i + 1 > maxlenth) {
                    maxlenth = j - i + 1;
                    left = i;
                    right = j;
                }
            }

        }
        return s.substr(left, right - left + 1);
    }
};

int main() {
    Solution sol;
    string s = "cbbd";
    string reslut=sol.longestPalindrome(s);
    cout << reslut << endl;
}