标签：span substring like substr als output 数组 复杂 size
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"
Solution 1:
两两对称验证。以每一个字符为中心，左右扩散来找回文串。注意奇数个aba和偶数个abba两种。时间复杂度是O(n2)
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int startIdx = 0; 5 int left = 0, right = 0; 6 int len = 0; 7 for(int i = 0; i < s.size() - 1; i++){ 8 if(s[i] == s[i + 1]){//偶数个palin 9 left = i; 10 right = i + 1; 11 searchPalin(s, left, right, startIdx, len); 12 } 13 left = right = i;//奇数个palin 14 searchPalin(s, left, right, startIdx, len); 15 } 16 if(len == 0) len = s.size();//in case like "a", 如果len是0， 则会return一个空串 17 return s.substr(startIdx, len); 18 } 19 20 void searchPalin(string s, int& left, int& right, int& startIdx, int& len){ 21 int step = 1;//左右延展判断的时候走的步数 22 while(left - step >= 0 && right + step < s.size()){ 23 if(s[left - step] != s[right + step]) break;//步数palin要break 24 ++step;//这里的step要比真正走的步数要大一，所以对于以后的计算len和startIdx的时候要小心 25 } 26 //if(len == 0) len = s.size(); 27 int wide = right - left + step * 2 - 1; 28 //startIdx = left - step + 1; 29 if(len < wide){ 30 len = wide; 31 startIdx = left - step + 1; 32 } 33 } 34 };
Solution 2: 动态规划DP
dp[i][j]:代表i到j之间是否为回文串。
dp[i, j] = 1 if i == j
= s[i] == s[j] if j = i + 1
= s[i] == s[j] && dp[i + 1][j - 1] if j > i + 1
这里有个有趣的现象就是如果我把下面的代码中的二维数组由int改为vector<vector<int> >后，就会超时，这说明int型的二维数组访问执行速度完爆std的vector啊，所以以后尽可能的还是用最原始的数据类型吧。
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int n = s.size(); 5 bool dp[n][n] = {false}; 6 int len = 0; 7 int left = 0, right = 0; 8 /* 9 for(int i = 0; i < n; i++){//初始化：单个字符是回文串 10 dp[i][i] = true; 11 } 12 for(int i = 0; i < n - 1; i++){ 13 dp[i][i + 1] = s[i] == s[i + 1];//初始化：j = i + 1的情况下 14 }*/ 15 //for(int j = 2; j < n; j++){//j > i + 1 16 //for(int i = 0; i < j - 1; i++){ 17 for(int j = 0; j < n; j++){//综合起来 18 for(int i = 0; i < j; i++){//i < j 所以dp[j][j]要写在最外层循环下面 19 //dp[i][j] = (s[i] == s[j]) && dp[i + 1][j - 1]; 20 dp[i][j] = (s[i] == s[j]) && (dp[i + 1][j - 1] || j - i <= 1); 21 if(dp[i][j] && j - i + 1 > len){ 22 len = j - i + 1; 23 left = i; 24 right = j; 25 } 26 //dp[j][j] = true; 27 } 28 dp[j][j] = true; 29 } 30 return s.substr(left, right - left + 1);//最后不是返回dp的值，而是substring 31 } 32 };
Solution 3: Manacher‘s Algorithms 马拉车算法 时间复杂度O(n)
5. Longest Palindromic Substring
标签：span substring like substr als output 数组 复杂 size
原文地址：http://www.cnblogs.com/93scarlett/p/6364053.html
