LeetCode最长回文子串（vector substr/dp）

20230125 完美默写
20230126 磕磕绊绊默写成功
20230129 更加磕磕绊绊的写对了
20230207 极度磕磕绊绊写对了
20230221 面向错误答案编程给怼对了
20230504

if(s[i] == s[j]){
    if(j - i < 3){
        dp[i][j] = true;
    }else
        dp[i][j] = dp[i + 1][j - 1];
}else{
    dp[i][j] = false;
}
if(dp[i][j] && ans < j - i + 1){
    ans = j - i + 1;
    l = i;
}

20230508 顺利通过
原题解

题目

约束

解法

解法一

#include <iostream>
#include <string>
#include <vector>

using namespace std;

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.size();
        if (n < 2) {
            return s;
        }

        int maxLen = 1;
        int begin = 0;
        // dp[i][j] 表示 s[i..j] 是否是回文串
        vector<vector<int>> dp(n, vector<int>(n));
        // 初始化：所有长度为 1 的子串都是回文串
        for (int i = 0; i < n; i++) {
            dp[i][i] = true;
        }
        // 递推开始
        // 先枚举子串长度
        for (int L = 2; L <= n; L++) {//长度是2~n
            // 枚举左边界，左边界的上限设置可以宽松一些
            for (int i = 0; i < n; i++) {
                // 由 L 和 i 可以确定右边界，即 j - i + 1 = L 得
                int j = L + i - 1;
                // 如果右边界越界，就可以退出当前循环
                if (j >= n) {//一定要判断越界的情况
                    break;
                }

                if (s[i] != s[j]) {
                    dp[i][j] = false;
                } else {
                    if (j - i < 3) {//包含两种情况，第一种是两位都一样第二种是三位中间任意两边一样。
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }

                // 只要 dp[i][j] == true 成立，就表示子串 s[i..j] 是回文，此时记录回文长度和起始位置
                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substr(begin, maxLen);
    }
};

1.字符串长度是2开始到n结束
2.要判断越界的情况下跳出循环

解法二

class Solution {
public:
    pair<int, int> expandAroundCenter(const string& s, int left, int right) {
        while (left >= 0 && right < s.size() && s[left] == s[right]) {
            --left;
            ++right;
        }
        return {left + 1, right - 1};
    }

    string longestPalindrome(string s) {
        int start = 0, end = 0;
        for (int i = 0; i < s.size(); ++i) {
            auto [left1, right1] = expandAroundCenter(s, i, i);
            auto [left2, right2] = expandAroundCenter(s, i, i + 1);
            if (right1 - left1 > end - start) {
                start = left1;
                end = right1;
            }
            if (right2 - left2 > end - start) {
                start = left2;
                end = right2;
            }
        }
        return s.substr(start, end - start + 1);
    }
};

解法三

class Solution {
public:
    int expand(const string& s, int left, int right) {
        while (left >= 0 && right < s.size() && s[left] == s[right]) {
            --left;
            ++right;
        }
        return (right - left - 2) / 2;
    }

    string longestPalindrome(string s) {
        int start = 0, end = -1;
        string t = "#";
        for (char c: s) {
            t += c;
            t += '#';
        }
        t += '#';
        s = t;

        vector<int> arm_len;
        int right = -1, j = -1;
        for (int i = 0; i < s.size(); ++i) {
            int cur_arm_len;
            if (right >= i) {
                int i_sym = j * 2 - i;
                int min_arm_len = min(arm_len[i_sym], right - i);
                cur_arm_len = expand(s, i - min_arm_len, i + min_arm_len);
            } else {
                cur_arm_len = expand(s, i, i);
            }
            arm_len.push_back(cur_arm_len);
            if (i + cur_arm_len > right) {
                j = i;
                right = i + cur_arm_len;
            }
            if (cur_arm_len * 2 + 1 > end - start) {
                start = i - cur_arm_len;
                end = i + cur_arm_len;
            }
        }

        string ans;
        for (int i = start; i <= end; ++i) {
            if (s[i] != '#') {
                ans += s[i];
            }
        }
        return ans;
    }
};