leetcode-5. Longest Palindromic Substring
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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"给定一个字符串s,判断最长回文子串。
参考了点击打开链接博主的做法,博主介绍了好几种做法,我只实现了比较容易的一个。
从头到尾遍历字符串s,以每个字符或两个字符为中心(回文有奇数和偶数)向两侧每次扩展一个位置,判断左右两侧的字符是否相同,由此记录中心形成的最长回文。
对每个中心形成的最长回文记录最长的。
class Solution {public: string centerJudge(string s, int left, int right){ string str = ""; //以一个字母或两个相同字母做中心,向两侧每次延伸一个字母,判断是否想同 while(s[left] == s[right] && left >= 0 && right < s.size()){ left --; right ++; } str = s.substr(left + 1, right - left - 1);//substr(off, count)函数,off为起始位置,count为字符串长度 //cout<<left<<" "<<right << " " <<str<<endl; return str; } string longestPalindrome(string s) { int n = s.size(); string longest = ""; string ans = ""; int maxlength = 0; for(int i = 0; i < n; i ++){ longest = centerJudge(s, i, i);//一个字母为中心 if(longest.size() > maxlength){ maxlength = longest.size(); ans = longest; } longest = centerJudge(s, i, i + 1);//两个相同字母为中心 if(longest.size() > maxlength){ maxlength = longest.size(); ans = longest; } } return ans; }};
注:substr(off, count)函数,off为起始位置,count为字符串长度
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