题目：
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
思路：
1) 两个for循环，用map存访问过的字母，遇到相同的就停止，然后再从这个相同字母的下一个字母开始。超时了。
package string; import java.util.HashMap; public class LongestSubstringWithoutRepeatingCharacters { public int lengthOfLongestSubstring(String s) { int len; if (s == null || (len = s.length()) == 0) return 0; HashMap<Character, Integer> map = new HashMap<Character, Integer>(); int max = 0; for (int i = 0; i < len; ++i) { map.put(s.charAt(i), i); for (int j = i + 1; j < len; ++j) { if (!map.containsKey(s.charAt(j))) { map.put(s.charAt(j), j); } else { i = map.get(s.charAt(j)); break; } } int count = map.size(); if (count > max) { max = count; } map.clear(); } return max; } public static void main(String[] args) { // TODO Auto-generated method stub String s = "abcabcbb"; LongestSubstringWithoutRepeatingCharacters l = new LongestSubstringWithoutRepeatingCharacters(); System.out.println(l.lengthOfLongestSubstring(s)); } }
2）对方法一的简化，保持重复元素的上一个位置，往前挪动，不断计算
package string; import java.util.HashMap; public class LongestSubstringWithoutRepeatingCharacters { public int lengthOfLongestSubstring(String s) { int len; if (s == null || (len = s.length()) == 0) return 0; HashMap<Character, Integer> map = new HashMap<Character, Integer>(); int max = 0; int lastIndex = -1; for (int i = 0; i < len; ++i) { char c = s.charAt(i); if (map.containsKey(c) && lastIndex < map.get(c)) { lastIndex = map.get(c); } if (i - lastIndex > max) max = i - lastIndex; map.put(c, i); } return max; } public static void main(String[] args) { // TODO Auto-generated method stub String s = "abcabcbb"; LongestSubstringWithoutRepeatingCharacters l = new LongestSubstringWithoutRepeatingCharacters(); System.out.println(l.lengthOfLongestSubstring(s)); } }
LeetCode - Longest Substring Without Repeating Characters
原文：http://www.cnblogs.com/shuaiwhu/p/5029295.html