求一个字符串中的最长回文子串。

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

 1 public class Solution {
 2     public String longestPalindrome(String s) {
 3         int length = 0;
 4         String res = "";
 5         
 6         //要考虑回文串是abba和aba两种情况
 7         for (int i = 0; i < 2 * s.length() - 1; i++) {
 8             int p1 = i % 2 == 0 ? i / 2 - 1 : i / 2;
 9             int p2 = i % 2 == 0 ? i / 2 + 1 : i / 2 + 1;
10             int len = i % 2 == 0 ? 1 : 0;
11             while (p1 >= 0 && p2 < s.length() && s.charAt(p1) == s.charAt(p2)) {
12                 len += 2;
13                 p1--;
14                 p2++;
15             }
16             if (len > length) {
17                 length = len;
18                 res = s.substring(p1 + 1, p2);
19             }
20         }
21         return res;
22     }
23 }

给一个字符串，找出其中不包含重复字符的最长子串的长度。

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

 1 public class Solution {
 2     public int lengthOfLongestSubstring(String s) {
 3         Map<Character, Integer> map = new HashMap<Character, Integer>();
 4         int res = 0;
 5         int p1 = 0;
 6         int p2 = 0;
 7         
 8         for (; p2 < s.length(); p2++) {
 9             char c = s.charAt(p2);
10             if (map.containsKey(c) && p1 <= map.get(c)) {
11                 p1 = map.get(c) + 1;
12             } 
13             map.put(c, p2);
14             res = Math.max(res, p2 - p1 + 1);
15         }
16         return res;
17     }
18 }