LeetCode 5. Longest Palindromic Substring

https://leetcode.com/problems/longest-palindromic-substring/description/

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"

• 最长回文子串。字符串处理题。可以深搜，动规，不过还是这个算法比较简单明了，找到回文串然后向两边扩张。

//
//  main.cpp
//  LeetCode
//
//  Created by Hao on 2017/3/16.
//  Copyright © 2017年 Hao. All rights reserved.
//

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.empty()) return "";
        if (s.size() == 1)  return s;
        
        size_t start = 0, maxLen = 1;
        
        for (size_t i = 0; i < s.size(); ) {
            if (s.size() - i <= maxLen / 2) break;
            size_t j = i, k = i;
            
            // Duplicate characters
            while ((k < s.size() - 1) && (s[k] == s[k + 1])) k ++;
            
            // Move start pointer
            i = k + 1;
            
            // Expand
            while ((k < s.size() - 1) && (j > 0) && (s[k + 1] == s[j - 1])) {
                k ++;
                j --;
            }
            
            if (k - j + 1 > maxLen) {
                maxLen = k - j + 1;
                start = j;
            }
        }
        
        return s.substr(start, maxLen);
    }
};

int main ()
{
    Solution testSolution;
    string sTest[] = {"babad", "cbbd"};

    for (int i = 0; i < 2; i ++)
        cout << testSolution.longestPalindrome(sTest[i]) << endl;
    
    return 0;
}

---

• Python算法同上，注意slicing左闭右开。

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s:
            return ""
        
        if len( s ) == 1:
            return s
        
        n_s = len( s )
        start, maxlen = 0, 1
        i = 0
        
        while i < n_s:
            if n_s - i <= maxlen / 2:
                break;
                
            left = right = i
            
            # move right point as duplicate chars is the center of palindromic substring
            while ( right + 1 < n_s ) and ( s[ right ] == s[ right + 1 ] ):
                right += 1
            
            # start point of next loop
            i = right + 1
                        
            # expand left and right points
            while ( left > 0 ) and ( right + 1 < n_s ) and ( s[ left - 1 ] == s[ right + 1 ] ):
                left -= 1
                right += 1
                
            if right - left + 1 > maxlen:
                maxlen = right - left + 1
                start = left

        # pay attention to slicing
        return s[ start : start + maxlen ]