LeetCode 5. Longest Palindromic Substring
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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, and there exists one unique longest palindromic substring.
题目解析
找出最大的回文子串;
开始的想法是遍历每一个字符,把第i个字符作为中心,用两指针法判断它往两边延伸是否构成回文,注意奇数回文和偶数回文差别。
然后看了一个参考代码,可以有效降低复杂度,把一段重复的字符串要只计算一次。
第一次代码
class Solution {public: string longestPalindrome(string s) { int max_l=0,max_r=0; for(int i=0;i<s.size();i++) { int l=i,r=i+1; while(l>=0&&r<s.size()&&s[l]==s[r]) { l--; r++; } if(r-l-2>max_r-max_l) { max_r=r-1; max_l=l+1; } l=i,r=i+2; while(l>=0&&r<s.size()&&s[l]==s[r]) { l--; r++; } if(r-l-2>max_r-max_l) { max_r=r-1; max_l=l+1; } } return s.substr(max_l,max_r-max_l+1); }};
引用参考代码
string longestPalindrome(string s) { if (s.empty()) return ""; if (s.size() == 1) return s; int min_start = 0, max_len = 1; for (int i = 0; i < s.size();) { if (s.size() - i <= max_len / 2) break; int j = i, k = i; while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters. i = k+1; while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand. int new_len = k - j + 1; if (new_len > max_len) { min_start = j; max_len = new_len; } } return s.substr(min_start, max_len); }
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