题目描述

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

Example 1:

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

Example 2:

Input: "cbbd"
Output: "bb"

题目思路

题目是求出最长回文字符串，千万不要理解错题意啊

• 思路一、参考

class Solution {
public:
    int maxLen = 0;
    int start = 0;
    
    string longestPalindrome(string s) {
        int len = s.size();
        if(len < 2) return s;
        
        for(int i=0; i < len-1; i++){
            fab(s, len, i, i);      // 如果为奇数
            fab(s, len, i, i+1);    // 如果为偶数
        }    
        return s.substr(start, maxLen);
    }
    
    void fab(string& s, int& len, int i, int j){
        while(i >= 0 && j < len && s[i]==s[j]){
            i--;
            j++;
        }
        
        if(maxLen < j - i - 1){
            maxLen = j - i - 1;
            start = i+1;
        }
    }
};

• 思路二、

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

• 思路三、multimap做的，耗时太多，因为超时，测试用例通过 101/103

class Solution {
public:
    string longestPalindrome(string s) {
        int len = s.size();
        
        multimap<char, int> mp;
        int maxstr = 1;
        int start = 0;
        int tt = 0;
        
        // 判断是不是完全一样的字符串
        int result = 0;
        for(int i=0; i < len; i++){
            if(s[i] == s[0]){
                result += 1;
            }
        }
        if(result == len) return s;
        
        for(int i=0; i < len; i++){
            // 找到了
            if(mp.count(s[i]) > 0){ 
                auto tt = mp.find(s[i]);
                for(int k=0; k != mp.count(s[i]); k++,tt++){
                    // 构成回文字符串
                    if(fab(s, tt->second, i) == true){
                        if(maxstr < i - tt->second + 1){
                            maxstr = i - tt->second + 1;
                            start = tt->second;
                        }
                    }
                }
            }
            mp.insert(make_pair(s[i], i));
        }
        
        return s.substr(start, maxstr);
    }
    
    bool fab(string& s, int i, int j){
        while(i < j){
            if(s[i++] != s[j--]){
                return false;
            }
        }
        return true;
    }
};

总结展望