5. Longest Palindromic Substring
题面
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
找出给定数组中，最长的回文子串。最长输入尾1000。
样例
Example 1:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.Example 2:
Input: "cbbd" Output: "bb"Example 3:
Input: "abcd" Output: "a"Example 4:
Input: "eeeeeeeeeeeeeeeeeeeeee" Output: "eeeeeeeeeeeeeeeeeeeeee"
思路1：暴力，不可行。
判断每一个字串是否是回文串，找出最长。
事件复杂度：O(n3) (1000的输入， n3复杂度， 10003=109，会超1000ms）
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 string res = ""; 5 int len = s.length(); 6 if(len == 0 || len == 1) 7 return s; 8 9 for(int i=0; i<len; i++) 10 { 11 string tmpStr = ""; 12 tmpStr += s[i]; 13 int j = i+1; 14 for(; j<len; j++) 15 { 16 if(s[j] != s[i]) 17 tmpStr += s[j]; 18 else 19 { 20 tmpStr += s[j]; 21 if(judge(tmpStr) && tmpStr.length() > res.length()) 22 res = tmpStr; 23 } 24 } 25 } 26 if(res.length() == 0) 27 res += s[0]; 28 return res; 29 } 30 31 bool judge(string s) 32 { 33 int len = s.length(); 34 for(int k=0; k<=len/2; k++) 35 if(s[k] != s[len-1-k]) 36 return false; 37 return true; 38 } 39 };
思路2：DP
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int len = s.length(); 5 if(len < 2) 6 return s; 7 bool dp[len][len] = {0}; 8 int start = 0, span = 1; 9 10 for(int i=0; i<len; i++) 11 { 12 dp[i][i] = 1; 13 for(int j=0; j<i; j++) 14 { 15 dp[j][i] = (s[i]==s[j] && (i-j < 2 || dp[j+1][i-1])); 16 if(dp[j][i] && (i-j+1 > span)) 17 { 18 span = i-j+1; 19 start = j; 20 } 21 } 22 } 23 return s.substr(start, span); 24 } 25 };