5. Longest Palindromic Substring(unsolved)
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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”
解答:这道题还没看懂,得再看
class Solution {public: 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);}};
二刷时的做法:
class Solution {public: string longestPalindrome(string s) { if(s.empty()) return s; int left=0,right=0,index=0,len=0; for(int i=0;i<s.size();i++){ if(s[i]==s[i+1]){ left=i; right=i+1; search(s,left,right,index,len); } left=i; right=i; search(s,left,right,index,len); } if(len==0) len=s.size(); return s.substr(index,len); } void search(string& s,int left,int right,int &index,int &len){ int step=1; while((left-step)>=0&&(right+step)<s.size()){ if(s[left-step]!=s[right+step]) break; step++; } int wide=right-left+2*step-1; if(len<wide){ len=wide; index=left-step+1; } }};
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