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"
1. O(n3)复杂度做。双层for循环,每次重新调用这个substring是不是palindrome的方程。
2.O(n2)复杂度做。DP。boolean dp[i][j]定义为s.substring(i, j + 1)是不是回文串。状态转移方程是:dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1]);
细节:1.不用担心i+1, j-1超过范围,因为i,j能进for循环说明满足要求了,而且i比j小,所以i+1最大到j,j-1最小到i,都是合格的。 2.需要担心i+1, j-1之间出现交叉,避免方式就是加上j-i < 3的限制,当两者靠的足够近有交叉的风险的时候你就去人工认可。3.注意最后返回的字符串是s.substring(i, j+1)不是i,j。
3.中心枚举法。O(n2)。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), 空间复杂度比DP的O(n2)要好。
2.DP实现
class Solution { public String longestPalindrome(String s) { if (s == null || s.length() == 0) { return ""; } boolean[][] dp = new boolean[s.length()][s.length()]; int ll = Integer.MIN_VALUE; int[] index = new int[2]; // the problem is caused when i and j are too close and i + 1 may cross with j - 1, but if they are close enough we can // just say by ourself. i==j, true, i + 1 == j, true, i + 2 == j, true // dp[i][j] = s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1]; for (int i = s.length() - 1; i >= 0; i--) { for (int j = i; j < s.length(); j++) { dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 3 || dp[i + 1][j - 1]); if(dp[i][j] && j - i > ll) { ll = j - i; index[0] = i; index[1] = j; } } } return s.substring(index[0], index[1] + 1); } }
3.枚举法实现
public class Solution { public String longestPalindrome(String s) { if (s == null || s.length() == 0) { return ""; } int start = 0, len = 0, longest = 0; for (int i = 0; i < s.length(); i++) { len = findLongestPalindromeFrom(s, i, i); if (len > longest) { longest = len; start = i - len / 2; } len = findLongestPalindromeFrom(s, i, i + 1); if (len > longest) { longest = len; start = i - len / 2 + 1; } } return s.substring(start, start + longest); } private int findLongestPalindromeFrom(String s, int left, int right) { int len = 0; while (left >= 0 && right < s.length()) { if (s.charAt(left) != s.charAt(right)) { break; } len += left == right ? 1 : 2; left--; right++; } return len; } }