5. Longest Palindromic Substring 暴力 800ms -> dp 97ms -> 回归原始 3ms AC

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"Output: "bab"Note: "aba" is also a valid answer.

Example:

Input: "cbbd"Output: "bb"

方法一：暴力解法，其中几个break很重要，否则会导致TLE， 800+ms AC

class Solution {public:    string longestPalindrome(string s) {        int maxL = 0, rl = 0, rr = 0;        int n = s.size();        for(int i = 0; i < s.size(); i++){            for(int j = n - 1; j >=i; j--){                int left = i, right = j;                while(left <= right){                    if(s[left] == s[right]){                        left++;                        right--;                    }else{                        break;                    }                                    }                if(right < left){                    if(maxL < j - i + 1){                        maxL = j - i + 1;                        rl = i;                        rr = j;                        break;                    }                }                if(maxL >= n - i - 1) break;            }        }        return s.substr(rl, rr - rl + 1);    }};

方法2：方法1的改进，用bool dp[i][j]表示 substr(i, j - i + 1)是否为回文 97ms AC

public class Solution {    public String longestPalindrome(String s) {        if(s == null || s.length() == 0) {            return "";        }        int len = s.length();        boolean[][] dp = new boolean[len][len];        int start = 0;        int end = 0;        int max = 0;        for(int i = 0; i < s.length(); i++) {            for(int j = 0; j <= i; j++) {                if(s.charAt(i) == s.charAt(j) && (i - j <= 2 || dp[j+1][i-1])) {                    dp[j][i] = true;                }                if(dp[j][i] && max < i - j + 1) {                    max = i - j + 1;                    start = j;                    end = i;                }            }        }        return s.substring(start, end + 1);    }}

方法3：以第i个元素左右散开（其实最先想到的就是这个方法，但是不知道那时候是脑抽了还是怎么的，认为这个方法复杂度是O(n^3)，就pass了，浪费了很多时间，，想到方法1，郁闷，实际上这个的复杂度最坏的情况下也仅仅是O(n^2)，也最简单）3ms AC

class Solution {public:    std::string longestPalindrome(std::string s) {        if (s.size() < 2)            return s;        int len = s.size(), max_left = 0, max_len = 1, left, right;        for (int start = 0; start < len && len - start > max_len / 2;) {            left = right = start;            while (right < len - 1 && s[right + 1] == s[right])                ++right;            start = right + 1;            while (right < len - 1 && left > 0 && s[right + 1] == s[left - 1]) {                ++right;                --left;            }            if (max_len < right - left + 1) {                max_left = left;                max_len = right - left + 1;            }        }        return s.substr(max_left, max_len);    }};