Leetcode - 5. Longest Palindromic Substring
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题目
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.
Simple solution(超时) O(N^3)
思路很简单,就是双循环不停的遍历所有的子字符串,然后检查是否为回稳并且目前是否最长。
class Solution {public: string longestPalindrome(string s) { int begin=0,len=1; int max=0; string res=""; for(begin=0;begin<s.length();begin++){ for(len=1;len+begin<=s.length();len++){ // attention the "<=" if(max<len && isPalind(s.substr(begin,len))){ max = len; res = s.substr(begin,len); } } } //cout<<isPalind(s); return res; } bool isPalind(string s){ int begin=0,end = s.length()-1; for(;begin<end;begin++,end--){ if(s[begin]!=s[end]){ return false; } } return true; }};
O(N^2)
思路:
对于每个子串的中心(可以是一个字符,或者是两个字符的间隙,比如串abc,中心可以是a,b,c,或者是ab的间隙,bc的间隙)往两边同时进行扫描,直到不是回文串为止。假设字符串的长度为n,那么中心的个数为2*n-1(字符作为中心有n个,间隙有n-1个)。对于每个中心往两边扫描的复杂度为O(n),所以时间复杂度为O((2*n-1)*n)=O(n^2),空间复杂度为O(1)
Note :
- std::string.substr(begin_idx,len)
- palindSubstr( )中的脚标: 由于left和right在结束时又自增(或自减),所以起始坐标是left+1;长度为(right-1)-(left+1)+1;
class Solution {public: string longestPalindrome(string s) { int left=0,right=0; int max=0; string res=""; for(int i=0;i<2*s.length()-1;i++){ left=i/2; //字符串aba~“a#b#a”,当断点为字符串间隔,right=left+1 right=(i+1)/2; string tmp = palindSubstr(s,left,right); if(max<tmp.length()){ res = tmp; max = tmp.length(); } } return res; } string palindSubstr(string s,int left, int right){ while(left>=0 && right<s.length() && s[left]==s[right]){ left--; right++; } return s.substr(left+1,right-left-1); }};
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